Full motivation. We claim that the only functions that work are $f(x)\equiv \frac{1}{x}$ and $f(x)\equiv 1$. These obviously work, we now show they are the only solutions. Assume from this point that $f$ is non-constant, as the only constant solution is obviously $f(x)\equiv 1$. Case 1: $f$ is non-injective. Then, we can find $a>b$ such that $f(a)=f(b)$. Let $a-b=c$. From $P(a,y)$, $P(b,y)$, we get $f(a+y)=f(b+y)$, hence $f(y)=f(y+c)$ for all $y>N$ for some constant $N$. This means $f(y)=f(y+kc)$ for any $k\in \mathbb{N}$ and $y>N$. If $f(x)>1$, then taking $k$ sufficiently large so that $1+f(x)(\frac{x+kc-1}{f(x)-1})>N$ and $\frac{x+kc-1}{f(x)-1}>0$, from $P(x,\frac{x+kc-1}{f(x)-1})$ we get $f(f(x))=1$. (This reduces to $x+y+kc=1+yf(x)$ and $y>0$.) Similarly, if $f(x)<1$, then taking $k$ sufficiently large so that $x+\frac{1+kc-x}{1-f(x)}>N$ and $\frac{1+kc-x}{1-f(x)}>0$, from $P(x, \frac{1+kc-x}{1-f(x)})$ we get $f(f(x))=1$. (This reduces to $x+y=1+yf(x)+kc$ and $y>0$.) Thus, either $f(x)=1$ or $f(f(x))=1$ for all $x$. This directly implies $f(f(1))=1$. $P(f(1),1)$ gives $f(1+f(1))f(1)=f(2)$ while $P(1,1)$ gives $f(2)=f(1+f(1))$, so $f(1)=1$. This in fact means $f(f(x))=1$ for all $x$, hence $f(x+y)=f(1+yf(x))$ for all $x,y$. Since $f$ is non-constant, suppose $f(t)\ne 1$. Then $f(t+y)=f(1+yf(t))$. From above, $f(t+y+z)=f(1+yf(t)+z)$. Take $d>0$. If $f(t)>1$, take $k$ sufficiently large so that $\frac{t+kc-1}{f(t)-1}>0$, taking $y=\frac{t+kc+d-1}{f(t)-1}$ in above gives $f(t+y+kc+d+z)=f(t+y+z)$. Since $z>N$ we have $f(t+y+d+z)=f(t+y+z)$, so for any $d>0$, $f(z)=f(z+d)$ for $z>N_d$ for some constant $N_d$. While $N_d>N$, then $f(z-c)=f(z-c+d)$, so we can replace $N_d$ with $N_d-c$. Thus, $f(z)=f(z+d)$ for any $z>N$ and $d>0$, which means $f$ is constant after some point. Suppose $f(x)=T$ for $x>N$. $P(x,y)$ for sufficiently large $x$ gives $T\cdot 1=f(1+yT)$, so $f(x)=T$ for $x>1$. Now $P(x,x): f(2x)=f(1+xf(x))=T$, so $f\equiv T$, a contradiction to $f$ being non-constant. If $f(t)<1$, a similar logic applies. Case 2: $f$ is injective. Note that if $f(x)-1$ and $x-1$ have same signs and $x\ne1$, $f(x)\ne 1$, then from $P(x,\frac{x-1}{f(x)-1})$, $f(f(x))=1$. Using this in the original equation with injectivity we get $x+y=1+yf(x)$, which means $x=1$ contradiction. Hence $f(x)-1$ and $x-1$ have different signs unless $x=1$ or $f(x)=1$. Thus, for any $x<1$ and $f(x)\ne 1$, $f(x)>1$, so $f(x+1-\frac{1}{f(x)})=1$, so $x+1-\frac{1}{f(x)}=k$ or $f(x)=\frac{1}{x+1-k}$ for all $x\ne k$, $x<1$. Notice that if $k>1$, then we have some $f(x)<0$, impossible. If $k>1$, then we can find some $x<1$ such that $f(x)=\frac{1}{x+1-k}<1$, contradiction. Hence $k=1$ and $f(x)=\frac{1}{x}$ for $x<1$. Note $f(k)=1$ means $f(1)=1$, so $f(x)=\frac{1}{x}$ for all $x\le 1$. $P(\frac{1}{x},\frac{1}{x})$ for $x\ge 2$ gives $f(x)=\frac{2f(2)}{x}$ for $x\ge 2$. Taking $x,\frac{y}{x}\to \infty$, $P(x,y): \frac{2f(2)}{x+y}\cdot \frac{x}{2f(2)}=f(1+\frac{y2f(2)}{x})=\frac{2f(2)}{1+\frac{y2f(2)}{x}}$ which gives $f(2)=\frac{1}{2}$. Hence $f(x)=\frac{1}{x}$ for $x\ge 2$ and $x\le 1$. Taking $x\to \infty$, $\frac{1}{x+y}\cdot x=f(1+\frac{y}{x})$, which means $f(x)=\frac{1}{x}$ for $x>1$. In conclusion, $f(x)=\frac{1}{x}$ for all $x>0$, which is the only non-constant function that works.

The answers are $f(x)=\frac{1}{x} \ \forall x \in \mathbb{R^+}$ and $f(x)=1 \ \forall x \in \mathbb{R^+}$, both can be checked to work. Let $P(x,y)$ denote the assertion. $P(x,yf(z)) \implies f(x+yf(z))f(f(x))=f(1+yf(x)f(z))=f(z+yf(x))f(f(z))$. $(\star)$ Claim: If $x>z$ and $f(x)>f(z)$, then $f(f(x))=f(f(z))$. Proof: Plugging $y=\frac{x-z}{f(x)-f(z)}$ in $(\star)$ gives the conclusion. $\blacksquare$ Case 1: $f$ is injective. If there exists $x$ such that $f(x)>1$, then $P(x,\frac{f(x)-1}{f(x)})$ gives that there exists $\alpha>0$ such that $f(\alpha)=1$. If $f(x)<1$ for all $x \in \mathbb{R^+}$, then for $x \in (0,1)$, $P(x,\frac{1-x}{1-f(x)})$ gives that there exists $\alpha>0$ such that $f(\alpha)=1$, so in either case, $1 \in \text{Img}(f)$. Now, for any $x \in (0,\alpha)$, $P(x,\alpha-x)$ gives $f(f(x))=f(1+(\alpha-x)f(x)) \implies f(x)=\frac{1}{x+1-\alpha}$ for all $x \in (0,\alpha)$, in particular, $\alpha\leq 1$. So, if $x,z \in \mathbb{R^+}$ such that $x < f(z)$, then plugging $y=\frac{f(z)-x}{f(z)}$ in $(\star)$ and using injectivity, we get $f(x)=z+\frac{(f(z)-x)f(x)}{f(z)}$. Plug $z=\alpha$, we get $f(x)=\frac{\alpha}{x}$ for all $x<1$, but since $f(x)=\frac{1}{x+1-\alpha}$ for all $x \in (0,\alpha)$, we get $\frac{1}{x+1-\alpha}=\frac{\alpha}{x} \implies \alpha = 1$. So $f(x)=\frac{1}{x}$ for all $x\leq 1$. Now let $x>1$ be any positive real, choose $z<1$ such that $x < \frac{1}{z}=f(z)$, then $f(x)=z+(1-zx)f(x) \implies f(x)=\frac{1}{x}$ for all $x>1$, so $f(x)=\frac{1}{x}$ for all $x \in \mathbb{R^+}$. Case 2: $f$ is not injective. Claim 1: If $a \neq b$ and $f(a)=f(b)$, then $f(f(u))=f(f(v))$ for all $u,v>\min{(a,b)}$. Proof: WLOG let $a < b$ and $f(a)=f(b)$. By $P(a,y)$ and $P(b,y)$ we obtain $f(y)=f(y+b-a)$ for all $y>a$. Let $u,v>a$. If $f(u)=f(v)$ then $f(f(u))=f(f(v))$. If $f(u) \neq f(v)$, assume WLOG that $f(u)>f(v)$. Take $k \in \mathbb{N}$ such that $u+k(b-a)>v$. Then by Claim, we have $f(f(u))=f(f(u+k(b-a)))=f(f(v))$, so done. $\blacksquare$ Claim 2: There exists $r$ such that $f(x)=f(y)$ for all $x,y>r$. Proof: Let $a<b$ and $f(a)=f(b)$. By Claim 1, we know $f(f(x))=d$ for all $x>a$ for some constant $d$. By $P(a,y)$ and $P(b,y)$ we also know $f(y)=f(y+b-a)$ for all $y>a$. We consider two cases. Case 2.1: There exists $\beta > a$ such that $f(\beta)>1$. In this case, $P(x,\frac{f(\beta)-1}{f(x)})$ for $x>a$ implies $f(x+\frac{f(\beta)-1}{f(x)})=1$ for all $x>a$. Now let $t>\max{(a,1)}$ be such that $f(t)=1$. Then, $f(1)=f(f(t))=f(f(\beta))$, and since $f(\beta)>1$, by Claim 1, we get $f(f(x))=f(f(y))$ for all $x,y>1$ and $P(1,y)$ and $P(f(\beta),y)$ gives $f(y)=f(y+c)$ for all $y>1$ for some constant $c$. Now let $x>1$, $P(x,y)$ and $P(x,y+c)$ implies $f(1+yf(x))=f(1+yf(x)+cf(x))$. Since $1+yf(x)$ can take any value $>1$, we get $f(y)=f(y+cf(x))$ for all $x,y>1$. In $(\star)$, if we choose $x,z>1$ and $y=c$, then $f(x)=f(x+cf(z))=f(z+cf(x))=f(z)$ for all $x,z>1$, thus claim is proven in this case. Case 2.2: $f(\beta) \leq 1$ for all $\beta>1$. Now let $k,\ell>a$ such that $f(k) \neq f(\ell)$. (This exists otherwise we are done) Then $f(f(k))=f(f(\ell))$, by Claim 1, $f(f(x))=f(f(y))$ for all $x>1$, also $P(f(k),y)$ and $P(f(\ell),y)$ gives that $f(y)=f(y+c)$ for all $y>1$ for some constant $c$. Now we can finish as in Case 2.1. $\blacksquare$ Now let $f(x)=\gamma$ for all $x>r$. Choosing $x>r$, $y>\frac{r}{\gamma}$ in $P(x,y)$, we obtain that $f(\gamma)=1$. Now choosing $x>r$ and $y=\frac{z-1}{\gamma}$ for $z>1$ in $P(x,y)$ we obtain that $f(z)=\gamma$ for all $z>1$. So, we can rewrite $P(x,y)$ as $f(y)f(f(x))=\gamma$ for all $y>x$. Choosing $y>\max{(x,r)}$ in $P(x,y)$ we get $f(f(x))=1$ for all $x \in \mathbb{R^+}$. So $f(y)=\gamma$ for all $y>0$, and plugging in, we get $\gamma=1$, so $f(x)=1$ for all $x \in \mathbb{R^+}$ is the only solution in this case. $\square$

Let $P(x,y)$ denote the assertion that $$f(x+y)f(f(x)) = f(1+yf(x))$$ Case 1: $f$ is not injective. There exists $a<b$ such that $f(a)=f(b)$ $P(a,y), P(b,y)$ yields $f(a+y)=f(b+y) \forall y>0$. Let $T=b-a$, then $f(x)=f(x+T)$ for all $x\ge a$. MAKE STUFF EQUAL: $x+y\equiv 1+yf(x) (\bmod\; T)$. This is doable for any $x$, so $f(f(x))\equiv 1$, and $$f(x+y)=f(1+yf(x))$$ Plugging $x=f(z)$ into that equation gives $f(f(z)+y)=f(1+y)$ for any $z,y$. Plugging $y=1$ gives $f(x+1)=f(1+f(x))=f(1+1)=f(2)$, so $f(x)= f(2)$ for all $x>1$. Now plugging any $x,y$ back into $f(x+y)=f(1+yf(x))=f(2)$ yields $f$ is constant, so $f$ is a constant function. Easy to check $f\equiv 1$. Case 2: $f$ is injective. MAKE STUFF EQUAL: fix $x$. If $x+y=1+yf(x)$ for some $y>0$, then $f(f(x))=1$ and $f(x+y)=f(1+yf(x))$ for all $y>0$. This contradicts $f$'s injectivity. Therefore, $x+y\ne 1+yf(x)$ for any $y>0$. In other words, if $x>1$, then $f(x)\le 1$, and if $x<1$ then $f(x)\ge 1$ MAKE STUFF EQUAL AGAIN: $f(x)=1+yf(x)$. Such $y>0$ exists as long as $f(x)>1$. Therefore, if $f(x)>1$ for some $x$, then $P(x, \frac{f(x)-1}{f(x)})$ gives $f(x+\frac{f(x)-1}{f(x)})=1$. Since $f$ is injective, there exists exactly one number $C$ such that $f(C)=1$. We have $C=x+\frac{f(x)-1}{f(x)}$, which means that $f(x) = \frac{1}{x+1-C}$ when $f(x)>1$. Claim: $C=1$ Proof: Assume not. $P(C,y)$ yields $$f(C+y)f(1)=f(1+y)$$ Case 1: $C>1$. Then $f(t)=f(1)f(t+(C-1))$ for all $t>1$. In particular, $f(t)=f(1)^k f(t+k(C-1))$. Pick $u$ such that $f(u)<1$. Suppose $y_1,y_2$ satisfy $u+y_1-1-y_1f(u)=k(C-1), u+y_2-1-y_2f(u)=(k+1)(C-1)$. Then $P(u,y_1), P(u,y_2)$ gives $f(f(u)) = f(1)^k = f(1)^{k+1}$, which implies that $f(1)=1$. This implies $C=1$. The case where $C<1$ is similar. Therefore, $f(x)\equiv \frac 1x$ whenever $f(x)>1$. We know that $x<1 \to f(x)\ge 1$ and $f(x)\ne f(1)=1$, so $f(x)=\frac 1x$ for all $x<1$. Now, pluggin $x,y$ such that $x+y<1$ gives $$\frac{1}{x+y} f(1/x) = f(1+y/x) = f\left( \frac{x+y}{x} \right)$$ For any $u>v>1$, there exists $x,y$ such that $1/x=u$ and $\frac{x+y}{x}=v$. This readily implies that if $g$ is $f$ restricted to $\mathbb{R}_{>1}$, then $xg(x)$ is constant, i.e. $xg(x)=V$ for some constant $V$. It's easy to prove $V=1$.

$\textbf{Answer:}$ $f \equiv 1, \frac{1}{x}$. It is easy to verify $f \equiv 1$, and for $\frac{1}{x}$ \[\text{LHS} = \frac{1}{x+y}f\left(\frac{1}{x}\right) = \frac{x}{x+y}\]\[\text{RHS} = f\left(1+\frac{y}{x}\right) = f\left(\frac{x+y}{x}\right) = \frac{x}{x+y}\]Hence verified to work. Let $P(x,y)$ be the assertion above. $\textbf{Case 1:}$ $f$ is not injective. We can assume we have $a \neq b, f(a) = f(b)$, and considering $P(x,a)$, $P(x,b)$ we get that \[f(x+a) = f(x+b)\]Hence we have $f$ is pseudo-periodic, say with value of $p$, i.e $f(x+p) = f(x)$ for $x \geq N$. Let us consider the value of $1+yf(x)-x-y$ with $x$ to be fixed. If we have that $f(x) \ne 1$, we can adjust the value of $y$ such that \[1+y(f(x)-1)-x = np\]where $n \in \mathbb{Z}$. But since $f$ is pseudo-periodic, notice taking $y \mapsto y \pm \frac{p}{f(x)-1}$ works for the above and hence indefinitely repeating the procedure will guarantee us to have $x+y, 1+yf(x) \geq N$, ultimately giving us $f(x+y) = f(1+yf(x))$. Hence we get for every $x$ that $f(x) = 1$ or $f(f(x)) = 1$. $\textbf{Sub Case 1:}$ $f(1) \ne 1$ and $f(f(1)) = 1$. Notice taking $P(1,y)$ gives us $f(1+y) = f(1+yf(1)) = f(1+yc)$ where $c \ne 1$. Now then notice in the original equation, taking $y \mapsto cy$, we get that \[f(x+y) = f(x+yc)\]Now but notice going back to the start of case $1$, we have if $a \ne b, f(a) = f(b)$, then $|a-b|$ is a pseudo-period value. But since we have $f(x+y) = f(x+yc) \implies |y(1-c)|$ is a pseudo-period value for all $y$ with no lower bound on the pseudo-period bound due to the $x, y$, which means we have all arbitrarily small $x \in \mathbb{R^+}$ as a pseudo-period value, giving $f(x) \equiv C$, which easily subbing gives finally $f(x) \equiv 1$. $\textbf{Sub Case 2:}$ $f(1) = 1$. This would mean that $f(f(x)) = 1$ for all $x \in \mathbb{R^+}$. This gives us that our original equation is \[f(x+y) = f(1+yf(x))\]Now taking $y \mapsto f(y)$, and considering symmetry, we get \[f(x+f(y)) = f(y+f(x))\]Taking $y \mapsto f(y)$ again, we get \[f(x+1) = f(f(x) + f(y))\]and again considering symmetry we get that $f(x+1) = f(y+1), \forall x,y \in \mathbb{R^+}$. Hence we get $f(x) = C, \forall x > 1$, and obviously returning back, we get that $f(x+y) = C \iff f(x) = C, \forall x \in \mathbb{R^+}$. Knowing that $f(f(x)) = 1$, we get that $f(x) = 1$ for all $x \in \mathbb{R^+}$. $\textbf{Case 2:}$ $f$ is injective. $\textbf{Claim:}$ There exists a $c$ such that $f(c) = 1$. $\textit{Proof}$ If for some $x$, we have that $f(x) > 1$, notice we can just take $P(x, \frac{f(x)-1}{f(x)})$ and we are done. Hence if we assume $f(x) < 1$ for all $x$, then notice if we consider the equation as so \[f(x+y) = \frac{f(1+yf(x))}{f(f(x))}\]Let us consider $x>1$ to be fixed and a $y' = \frac{x+y-1}{f(x)}$. Notice this would make $f(1+y'f(x)) = f(x+y)$. \[f(x+y') = \frac{f(1+y'f(x))}{f(f(x))} = \frac{f(1+yf(x))}{f(f(x))^2}\]Notice we can continue this process indefinitely and the value increases at a constant multiplicative rate as $f(f(x)) < 1$ is fixed. Hence at some point we will reach a $f(x) > 1$. $\blacksquare$ Thus now we can consider $P(x, c-x)$ to get \[f(f(x)) = f(1+(c-x)f(x))\]\[f(x) = 1+(c-x)f(c)\]\[f(x) = \frac{1}{1-c+x}\]for all $x \leq c$. Notice $c \leq 1$ as if its not, we can set $x$ to be arbitrarily small to get $f(x)$ is negative which is a contradiction. $\textbf{Claim:}$ If $x > 1$, we have $f(x) < 1$ and vice versa unless $f(x) = c$. $\textit{Proof}$ Notice if $x > 1, f(x) > 1$, we get by $P(x, \frac{x-1}{f(x)-1})$ that $f(f(x)) = 1 \implies f(x) = c$ contradiction. Easy to see for $x < 1$ by symmetry. $\blacksquare$ Now if $f(x) > 1$, as before taking $P(x, \frac{f(x)-1}{f(x)})$, gives us that $f(x+ \frac{f(x)-1}{f(x)}) = 1 \implies x+\frac{f(x)-1}{f(x)}=c \implies f(x) = \frac{1}{1-c+x}$. Hence we have $f(x) > 1$ if and only if $x \in (0, c]$. But then considering for $x$'s in the interval $(c, 1]$ gives us a contradiction with the claim. Thus this gives us that $c = 1 \implies f(x) = \frac{1}{x}, x \leq 1$. Hence now we finish easily off. Our proven claim now gets adjusted to $f(x) > 1 \iff x < 1$. Now if we let $x< 1$ in original equation, we get \[f(x+y)f\left(\frac{1}{x}\right) = f\left(\frac{x+y}{x}\right)\]Now if we consider $x+y > 1$, notice since we can attain any pair of values $x+y, \frac{1}{x}$ both greater than 1, we get $f$ is multiplicative over values greater than $1$. Now then in the original equation, considering $x > 1$, we get \[f(x+y)\frac{1}{f(x)} = f(1+yf(x))\]\[\implies f(x+y) = f(1+yf(x))f(x)\]\[\implies f(x+y) = f(x+xyf(x))\]\[\implies x+y = x+xyf(x)\]\[\implies f(x) = \frac{1}{x}, x > 1\]Hence this gives us $f(x) \equiv \frac{1}{x}$ for all $x \in \mathbb{R^+}$.