The only functions that work are $f(x)=L$ for some $L \geq 1$ and $f(x)=\frac{x+1}{x}$. Lemma 1: There exist $C,\varepsilon>0$ such that for all $x<\varepsilon$ we have $f(x)<\frac{C}{x}$. ProofCase 1: $f$ is bounded in $[1,\infty]$ Substitute $y=\frac{f(x)}{z}$ and multiply by $z$ to get: $$zf(z)-zf \left (\frac{f(\frac{f(x)}{z})} {x}\right) = xf(x)\left((f(x+1)-f\left(\frac{f(x)}{z}+1\right)\right)$$We can take $z \to 0$ to get: $$zf(z)-O(1) = xf(x)\left((f(x+1)-O(1)\right) \implies zf(z) = O(1)$$Which is what we wanted. Case 2: $f$ is not bounded in $[1,\infty]$ In particular, we can find big enough $x_0$ such that $f(x_0+1)>f(1)$. It means that for small enough $y$ the RHS is positive, which means that the LHS is also positive, i.e. $f\left(\frac{f(x_0)}{y}\right)>f\left(\frac{f(y)}{x_0}\right)$ for small enough $y$. Notice that $\frac{f(x_0)}{y},\frac{f(y)}{x_0}$ are both continuous, which means that the sign of $\frac{f(x_0)}{y}-\frac{f(y)}{x_0}$ is constant for small enough $x$, otherwise we would have by the intermediate value theorem small $y$ for which $\frac{f(x_0)}{y}=\frac{f(y)}{x_0}$, which is a contradiction to the positivity of the LHS in the original equation. We want to find the sign of $\frac{f(x_0)}{y}-\frac{f(y)}{x_0}$. Assume for contradiction that $\frac{f(x_0)}{y}<\frac{f(y)}{x_0}$ for all small enough $y$. In particular. $\frac{f(y)}{x_0} \to \infty$ when $y \to 0$, and because $f$ is continuous and unbounded in $[1,\infty]$ we can take $\frac{f(y)}{x_0}$ to be a global maximum of $f$ in the segment $[1,\frac{f(y)}{x_0}]$, which means that $$f\left(\frac{f(x_0)}{y}\right)<f\left(\frac{f(y)}{x_0}\right)$$Which is a contradiction. Therefore, $\frac{f(x_0)}{y}>\frac{f(y)}{x_0}$ for all small enough $y$, which proves the lemma. We now want to understand how $f(x)$ behaves when $x \to \infty$. There are $3$ cases: Case 1: $f$ has at least $2$ partial limits at $x\to \infty$ solutionAs $f$ is continuous, the set of partial limits is a segment (which might contain $\infty$), and every partial limit $L$ that isn't in the edge of that segment appears as $f(x)$ for values of $x$ as big as we want. Take two such partial limits $L_1>L_2$, fix a big value of $x_0$ such that $f(x_0)=L_1$ and take a sequence $y_n\to \infty$ such that $f(y_n)=L_2$. Substitute $x_0,y_m$ in the fe and multiply it by $\frac{f(x_0)}{y}$ to get: $$\frac{f(x_0)}{y_n}f\left(\frac{f(x_0)}{y_n}\right) > x_0f(x_0)\left(f(x_0+1)-f(y_n+1)\right)\ \to x_0f(x_0)(L_1-L_2)$$Notice that the LHS is bounded by $C$ by lemma 1, and therefore we get a contradiction by choosing $x_0> \frac {C}{L_1-L_2}$ (because we know that $f(x_0)>1$). Case 2: $lim_{x\to\infty}f(x) = \infty$ solutionLemma: $\frac{f(x)}{x}$ is unbounded in $[1,\infty]$. proofAssume for contradiction that $f(x)\leq cx$ for all $x\in[1,\infty]$. Then $f(f(x)) \leq c^2x$. Take $x\to \infty$ in the original equation and divide by $x$ to get: $$c^2 \geq \frac{f(f(x))}{x}>y\left(f(x+1)-f(y+1)\right)\ \to \infty$$Which is a contradiction. Notice that it means that $\frac{f(x)}{x}$ is unbounded in $[c,\infty]$ for all $c$ because of the continuity. By the proof of the lemma 1, we know that for all big enough $x$, for small enough $y$ (how small depends on $x$) we have $\frac{f(x)}{y}>\frac{f(y)}{x}$. Take some big $x_0$ and take $y \to 0$. Let $\varepsilon > 0$. Notice that $$0<f\left(\frac{f(x_0)}{y}\right)-f\left(\frac{f(y)}{x_0}\right) = x_0y\left(f(x_0+1)-f(y+1)\right)\ \to 0$$And in particular, for small enough $y$, the above expression is less than $\varepsilon$. We can take small $y$ such that $\frac{f(x_0)}{y}$ is a global maximum of $\frac{f(t)}{t}$ in the segment $[\frac{1}{x},\frac{f(x_0)}{y}]$. Let $a=f\left(\frac{f(x_0)}{y}\right),b=\frac{f(x_0)}{y},c=f\left(\frac{f(y)}{x_0}\right),d=\frac{f(y)}{x_0}$. Then by the maximality of $\frac{f(b)}{b}$ we have: $$\frac{a}{b} \geq \frac{c}{d}$$We can also assume that $ \frac{a}{b} \geq 1$, as it is the maximum value of $\frac{f(t)}{t}$ in the segment we are looking at and we know that $\frac{f(t)}{t}$ is unbounded in $t \to \infty$. Therefore: $$1\leq \frac{a}{b} \geq \frac{c}{d} \implies a-c \geq b-d$$(This is an easy inequality, I might add a proof for it for completeness later). Therefore: $$\varepsilon>f\left(\frac{f(x_0)}{y}\right)-f\left(\frac{f(y)}{x_0}\right)\geq \frac{f(x_0)}{y}-\frac{f(y)}{x_0} = \frac{x_0f(x_0)-yf(y)}{xy}$$and as $y$ is very close to $0$, we have in particular: $$\varepsilon>x_0f(x_0)-yf(y)$$But we know from lemma 1 that $yf(y)<C$, and $x_0f(x_0)$ can be as big as we want, which is a contradiction. Case 3: There exists some $L<\infty$ such that $lim_{x\to\infty}f(x) = L$ solutionSubstitute $y=\frac{f(x)}{z},z\to 0$ and multiply by $z$ to get: $$zf(z) \to xf(x)\left((f(x+1)-L\right)$$Therefore, when $z\to 0$, $zf(z) \to C$ for some $C$. If $C=0$, we get that $f(x+1)=L$ for all $x$, which means that $f$ is constant in $(1,\infty)$, and it's easy to check it means that $f$ must be constant in all of $(0,\infty)$. If $C\neq 0$, notice that when $z\to 0$ we have $f(z)\to\infty$. Substitute again $y=\frac{f(x)}{z}$ to get: $$f(z)-f \left (\frac{f(\frac{f(x)}{z})} {x}\right) = \frac{xf(x)}{z}\left((f(x+1)-f\left(\frac{f(x)}{z}+1\right)\right)$$By taking $x\to 0$ we get: $$f(z)-L = \frac{C}{z}(f(1)-L)$$We can substitute it in the original equation and see that the only solutions are $f(x)=L$ for some $L \geq 1$ and $f(x)=\frac{x+1}{x}$.

Let $xf(x) > x_0f(x_0)$ for all $0<x<\epsilon$ Since $f$ is unbounded on $[1,\infty]$, there exists $s>\max(f(\epsilon) / x_0, x_0)$ such that $f(s) \ge f(t)\forall t\in[x_0, s]$ Since $f(t)\to\infty$ as $t\to 0$ we know that $\exists t<\epsilon$ such that $f(t) = x_0 s$ We have that $\frac{f(x_0)}{t} < \frac{f(t)}{x_0} = s$ Putting it all together we have that $f\left(\frac{f(t)}{x_0}\right) = f(s) \ge f\left(\frac{f(x_0)}{t}\right)$ Which is a contradiction.

Let $xf(x) > x_0f(x_0)$ for all $0<x<\epsilon$ Since $f$ is unbounded on $[1,\infty]$, there exists $s>x_0$ such that $f(s) \ge f(t)\forall t\in[x_0, s]$ Since $f(t)\to\infty$ as $t\to 0$ we know that $\exists t<\epsilon$ such that $f(t) = x_0 s$ We have that $\frac{f(x_0)}{t} < \frac{f(t)}{x_0} = s$ Putting it all together we have that $f\left(\frac{f(t)}{x_0}\right) = f(s) \ge f\left(\frac{f(x_0)}{t}\right)$ Which is a contradiction.

The only functions that work are $f(x)=L$ for some $L \geq 1$ and $f(x)=\frac{x+1}{x}$. Lemma 1: There exist $C,\varepsilon>0$ such that for all $x<\varepsilon$ we have $f(x)<\frac{C}{x}$. ProofCase 1: $f$ is bounded in $[1,\infty]$ Substitute $y=\frac{f(x)}{z}$ and multiply by $z$ to get: $$zf(z)-zf \left (\frac{f(\frac{f(x)}{z})} {x}\right) = xf(x)\left((f(x+1)-f\left(\frac{f(x)}{z}+1\right)\right)$$We can take $z \to 0$ to get: $$zf(z)-O(1) = xf(x)\left((f(x+1)-O(1)\right) \implies zf(z) = O(1)$$Which is what we wanted. Case 2: $f$ is not bounded in $[1,\infty]$ In particular, we can find big enough $x_0$ such that $f(x_0+1)>f(1)$. It means that for small enough $y$ the RHS is positive, which means that the LHS is also positive, i.e. $f\left(\frac{f(x_0)}{y}\right)>f\left(\frac{f(y)}{x_0}\right)$ for small enough $y$. Notice that $\frac{f(x_0)}{y},\frac{f(y)}{x_0}$ are both continuous, which means that the sign of $\frac{f(x_0)}{y}-\frac{f(y)}{x_0}$ is constant for small enough $x$, otherwise we would have by the intermediate value theorem small $y$ for which $\frac{f(x_0)}{y}=\frac{f(y)}{x_0}$, which is a contradiction to the positivity of the LHS in the original equation. We want to find the sign of $\frac{f(x_0)}{y}-\frac{f(y)}{x_0}$. Assume for contradiction that $\frac{f(x_0)}{y}<\frac{f(y)}{x_0}$ for all small enough $y$. In particular. $\frac{f(y)}{x_0} \to \infty$ when $y \to 0$, and because $f$ is continuous and unbounded in $[1,\infty]$ we can take $\frac{f(y)}{x_0}$ to be a global maximum of $f$ in the segment $[1,\frac{f(y)}{x_0}]$, which means that $$f\left(\frac{f(x_0)}{y}\right)<f\left(\frac{f(y)}{x_0}\right)$$Which is a contradiction. Therefore, $\frac{f(x_0)}{y}>\frac{f(y)}{x_0}$ for all small enough $y$, which proves the lemma. We now want to understand how $f(x)$ behaves when $x \to \infty$. There are $3$ cases: Case 1: $f$ has at least $2$ partial limits at $x\to \infty$ solutionAs $f$ is continuous, the set of partial limits is a segment (which might contain $\infty$), and every partial limit $L$ that isn't in the edge of that segment appears as $f(x)$ for values of $x$ as big as we want. Take two such partial limits $L_1>L_2$, fix a big value of $x_0$ such that $f(x_0)=L_1$ and take a sequence $y_n\to \infty$ such that $f(y_n)=L_2$. Substitute $x_0,y_m$ in the fe and multiply it by $\frac{f(x_0)}{y}$ to get: $$\frac{f(x_0)}{y_n}f\left(\frac{f(x_0)}{y_n}\right) > x_0f(x_0)\left(f(x_0+1)-f(y_n+1)\right)\ \to x_0f(x_0)(L_1-L_2)$$Notice that the LHS is bounded by $C$ by lemma 1, and therefore we get a contradiction by choosing $x_0> \frac {C}{L_1-L_2}$ (because we know that $f(x_0)>1$). Case 2: $lim_{x\to\infty}f(x) = \infty$ solutionLemma: $\frac{f(x)}{x}$ is unbounded in $[1,\infty]$. proofAssume for contradiction that $f(x)\leq cx$ for all $x\in[1,\infty]$. Then $f(f(x)) \leq c^2x$. Take $x\to \infty$ in the original equation and divide by $x$ to get: $$c^2 \geq \frac{f(f(x))}{x}>y\left(f(x+1)-f(y+1)\right)\ \to \infty$$Which is a contradiction. Notice that it means that $\frac{f(x)}{x}$ is unbounded in $[c,\infty]$ for all $c$ because of the continuity. By the proof of the lemma 1, we know that for all big enough $x$, for small enough $y$ (how small depends on $x$) we have $\frac{f(x)}{y}>\frac{f(y)}{x}$. Take some big $x_0$ and take $y \to 0$. Let $\varepsilon > 0$. Notice that $$0<f\left(\frac{f(x_0)}{y}\right)-f\left(\frac{f(y)}{x_0}\right) = x_0y\left(f(x_0+1)-f(y+1)\right)\ \to 0$$And in particular, for small enough $y$, the above expression is less than $\varepsilon$. We can take small $y$ such that $\frac{f(x_0)}{y}$ is a global maximum of $\frac{f(t)}{t}$ in the segment $[\frac{1}{x},\frac{f(x_0)}{y}]$. Let $a=f\left(\frac{f(x_0)}{y}\right),b=\frac{f(x_0)}{y},c=f\left(\frac{f(y)}{x_0}\right),d=\frac{f(y)}{x_0}$. Then by the maximality of $\frac{f(b)}{b}$ we have: $$\frac{a}{b} \geq \frac{c}{d}$$We can also assume that $ \frac{a}{b} \geq 1$, as it is the maximum value of $\frac{f(t)}{t}$ in the segment we are looking at and we know that $\frac{f(t)}{t}$ is unbounded in $t \to \infty$. Therefore: $$1\leq \frac{a}{b} \geq \frac{c}{d} \implies a-c \geq b-d$$(This is an easy inequality, I might add a proof for it for completeness later). Therefore: $$\varepsilon>f\left(\frac{f(x_0)}{y}\right)-f\left(\frac{f(y)}{x_0}\right)\geq \frac{f(x_0)}{y}-\frac{f(y)}{x_0} = \frac{x_0f(x_0)-yf(y)}{xy}$$and as $y$ is very close to $0$, we have in particular: $$\varepsilon>x_0f(x_0)-yf(y)$$But we know from lemma 1 that $yf(y)<C$, and $x_0f(x_0)$ can be as big as we want, which is a contradiction. Case 3: There exists some $L<\infty$ such that $lim_{x\to\infty}f(x) = L$ solutionSubstitute $y=\frac{f(x)}{z},z\to 0$ and multiply by $z$ to get: $$zf(z) \to xf(x)\left((f(x+1)-L\right)$$Therefore, when $z\to 0$, $zf(z) \to C$ for some $C$. If $C=0$, we get that $f(x+1)=L$ for all $x$, which means that $f$ is constant in $(1,\infty)$, and it's easy to check it means that $f$ must be constant in all of $(0,\infty)$. If $C\neq 0$, notice that when $z\to 0$ we have $f(z)\to\infty$. Substitute again $y=\frac{f(x)}{z}$ to get: $$f(z)-f \left (\frac{f(\frac{f(x)}{z})} {x}\right) = \frac{xf(x)}{z}\left((f(x+1)-f\left(\frac{f(x)}{z}+1\right)\right)$$By taking $x\to 0$ we get: $$f(z)-L = \frac{C}{z}(f(1)-L)$$We can substitute it in the original equation and see that the only solutions are $f(x)=L$ for some $L \geq 1$ and $f(x)=\frac{x+1}{x}$.

The only functions that work are $f(x)=L$ for some $L \geq 1$ and $f(x)=\frac{x+1}{x}$. Lemma 1: There exist $C,\varepsilon>0$ such that for all $x<\varepsilon$ we have $f(x)<\frac{C}{x}$. ProofCase 1: $f$ is bounded in $[1,\infty]$ Substitute $y=\frac{f(x)}{z}$ and multiply by $z$ to get: $$zf(z)-zf \left (\frac{f(\frac{f(x)}{z})} {x}\right) = xf(x)\left((f(x+1)-f\left(\frac{f(x)}{z}+1\right)\right)$$We can take $z \to 0$ to get: $$zf(z)-O(1) = xf(x)\left((f(x+1)-O(1)\right) \implies zf(z) = O(1)$$Which is what we wanted. Case 2: $f$ is not bounded in $[1,\infty]$ In particular, we can find big enough $x_0$ such that $f(x_0+1)>f(1)$. It means that for small enough $y$ the RHS is positive, which means that the LHS is also positive, i.e. $f\left(\frac{f(x_0)}{y}\right)>f\left(\frac{f(y)}{x_0}\right)$ for small enough $y$. Notice that $\frac{f(x_0)}{y},\frac{f(y)}{x_0}$ are both continuous, which means that the sign of $\frac{f(x_0)}{y}-\frac{f(y)}{x_0}$ is constant for small enough $x$, otherwise we would have by the intermediate value theorem small $y$ for which $\frac{f(x_0)}{y}=\frac{f(y)}{x_0}$, which is a contradiction to the positivity of the LHS in the original equation. We want to find the sign of $\frac{f(x_0)}{y}-\frac{f(y)}{x_0}$. Assume for contradiction that $\frac{f(x_0)}{y}<\frac{f(y)}{x_0}$ for all small enough $y$. In particular. $\frac{f(y)}{x_0} \to \infty$ when $y \to 0$, and because $f$ is continuous and unbounded in $[1,\infty]$ we can take $\frac{f(y)}{x_0}$ to be a global maximum of $f$ in the segment $[1,\frac{f(y)}{x_0}]$, which means that $$f\left(\frac{f(x_0)}{y}\right)<f\left(\frac{f(y)}{x_0}\right)$$Which is a contradiction. Therefore, $\frac{f(x_0)}{y}>\frac{f(y)}{x_0}$ for all small enough $y$, which proves the lemma. We now want to understand how $f(x)$ behaves when $x \to \infty$. There are $3$ cases: Case 1: $f$ has at least $2$ partial limits at $x\to \infty$ solutionAs $f$ is continuous, the set of partial limits is a segment (which might contain $\infty$), and every partial limit $L$ that isn't in the edge of that segment appears as $f(x)$ for values of $x$ as big as we want. Take two such partial limits $L_1>L_2$, fix a big value of $x_0$ such that $f(x_0)=L_1$ and take a sequence $y_n\to \infty$ such that $f(y_n)=L_2$. Substitute $x_0,y_m$ in the fe and multiply it by $\frac{f(x_0)}{y}$ to get: $$\frac{f(x_0)}{y_n}f\left(\frac{f(x_0)}{y_n}\right) > x_0f(x_0)\left(f(x_0+1)-f(y_n+1)\right)\ \to x_0f(x_0)(L_1-L_2)$$Notice that the LHS is bounded by $C$ by lemma 1, and therefore we get a contradiction by choosing $x_0> \frac {C}{L_1-L_2}$ (because we know that $f(x_0)>1$). Case 2: $lim_{x\to\infty}f(x) = \infty$ solutionLemma: $\frac{f(x)}{x}$ is unbounded in $[1,\infty]$. proofAssume for contradiction that $f(x)\leq cx$ for all $x\in[1,\infty]$. Then $f(f(x)) \leq c^2x$. Take $x\to \infty$ in the original equation and divide by $x$ to get: $$c^2 \geq \frac{f(f(x))}{x}>y\left(f(x+1)-f(y+1)\right)\ \to \infty$$Which is a contradiction. Notice that it means that $\frac{f(x)}{x}$ is unbounded in $[c,\infty]$ for all $c$ because of the continuity. By the proof of the lemma 1, we know that for all big enough $x$, for small enough $y$ (how small depends on $x$) we have $\frac{f(x)}{y}>\frac{f(y)}{x}$. Take some big $x_0$ and take $y \to 0$. Let $\varepsilon > 0$. Notice that $$0<f\left(\frac{f(x_0)}{y}\right)-f\left(\frac{f(y)}{x_0}\right) = x_0y\left(f(x_0+1)-f(y+1)\right)\ \to 0$$And in particular, for small enough $y$, the above expression is less than $\varepsilon$. We can take small $y$ such that $\frac{f(x_0)}{y}$ is a global maximum of $\frac{f(t)}{t}$ in the segment $[\frac{1}{x},\frac{f(x_0)}{y}]$. Let $a=f\left(\frac{f(x_0)}{y}\right),b=\frac{f(x_0)}{y},c=f\left(\frac{f(y)}{x_0}\right),d=\frac{f(y)}{x_0}$. Then by the maximality of $\frac{f(b)}{b}$ we have: $$\frac{a}{b} \geq \frac{c}{d}$$We can also assume that $ \frac{a}{b} \geq 1$, as it is the maximum value of $\frac{f(t)}{t}$ in the segment we are looking at and we know that $\frac{f(t)}{t}$ is unbounded in $t \to \infty$. Therefore: $$1\leq \frac{a}{b} \geq \frac{c}{d} \implies a-c \geq b-d$$(This is an easy inequality, I might add a proof for it for completeness later). Therefore: $$\varepsilon>f\left(\frac{f(x_0)}{y}\right)-f\left(\frac{f(y)}{x_0}\right)\geq \frac{f(x_0)}{y}-\frac{f(y)}{x_0} = \frac{x_0f(x_0)-yf(y)}{xy}$$and as $y$ is very close to $0$, we have in particular: $$\varepsilon>x_0f(x_0)-yf(y)$$But we know from lemma 1 that $yf(y)<C$, and $x_0f(x_0)$ can be as big as we want, which is a contradiction. Case 3: There exists some $L<\infty$ such that $lim_{x\to\infty}f(x) = L$ solutionSubstitute $y=\frac{f(x)}{z},z\to 0$ and multiply by $z$ to get: $$zf(z) \to xf(x)\left((f(x+1)-L\right)$$Therefore, when $z\to 0$, $zf(z) \to C$ for some $C$. If $C=0$, we get that $f(x+1)=L$ for all $x$, which means that $f$ is constant in $(1,\infty)$, and it's easy to check it means that $f$ must be constant in all of $(0,\infty)$. If $C\neq 0$, notice that when $z\to 0$ we have $f(z)\to\infty$. Substitute again $y=\frac{f(x)}{z}$ to get: $$f(z)-f \left (\frac{f(\frac{f(x)}{z})} {x}\right) = \frac{xf(x)}{z}\left((f(x+1)-f\left(\frac{f(x)}{z}+1\right)\right)$$By taking $x\to 0$ we get: $$f(z)-L = \frac{C}{z}(f(1)-L)$$We can substitute it in the original equation and see that the only solutions are $f(x)=L$ for some $L \geq 1$ and $f(x)=\frac{x+1}{x}$.