The zero and identity function work. We claim these are the only ones. It's easy to see that $f(0)=0$. If $f(c)=0$ for $c\neq 0$, then plugging in $y=c$ implies $f\equiv 0$. So assume $f(x)=0\iff x=0$. Now suppose $f(a)\neq a$ for some $a\neq 0$. Putting $x=a$ and $y=\frac{a}{f(a)-a)}$ gives $$ \frac{f(a)}{f(a)-a} f\left(\frac{af(a)}{f(a)-a}\right) = \frac{a}{f(a)-a}f\left(\frac{af(a)}{f(a)-a}\right)$$Note that $f\left(\frac{af(a)}{f(a)-a}\right )\neq 0$ since $\frac{af(a)}{f(a)-a}\neq 0$. Hence the above implies $f(a)=a$, a clear contradiction. So only the claimed functions work.

solution snipped This was my solution: We can see that $f(x)\equiv 0$ és $f(x)\equiv x$ there are solutions. We will prove that there not other solutiuons. Let $a\in \mathbb{R}$ with $f(a)\ne a$. In the same time let $y=\frac{a}{f(a)-a)}$ so $f(a)=\frac{a(y+1)}{y}\Leftrightarrow yf(a)=a(y+1)$ and replacing in the original equation, we get $(y+1)f(a(y+1))=yf(a(y+1)),\text{ }\forall y\in {{\mathbb{R}}^{*}}$ vagyis $y+1=y,\text{ }\forall y\in {{\mathbb{R}}^{*}}$ and this is contradiction.

Let $P(x,y)$ be the assertion. clearly, if $f(x)$ is a constant then $f(x)=0$ Consider the case $f(x)\neq0 \forall x\in\mathbb{R}$ $P(x,0)\Rightarrow f(0)=0$ Suppose that there exist $c\neq0$ with$f(c)=0$ $P(c,x)\Rightarrow f(c(x+1))=0 $ wich gives $f(x)=0$ a contradiction. So $f(c)=0\Leftrightarrow c=0$ Now suppose that $f(x)\neq{x}$ $P\left(x,\frac{1}{\frac{f(x)}{x}-1}\right )\Rightarrow f\left(\frac{f(x)}{\frac{f(x)}{x}-1}\right )=0 $ implies $f(x)=0$ a contradiction! so $f(x)=x$ easy to check or $f(x)=0$ are the only solutions and we are done ^ω^

Cycle very nice solution

Denote the assertion by $P(x,y).$ Clearly $f(0)=0.$ If $f(k)=0$ for some $k\neq 0$ then $P(x,k)$ gives $f(x)=0$ this works. If $f$ is not identically zero then $f$ is injective at zero. Assume $f(x)\neq x.$ Then $P(x,\tfrac{x}{f(x)-x})$ gives $f(x)=0$ impossible. Check that $f(x)\equiv x$ works.

Let $P(x,y)$ denote the given assertion. $P(0,0): f(0) =0$. If $f(k)=0$ for $k\ne 0$, $P(k,x): 0=xf(k(x+1))$, so $\boxed{f\equiv 0}$, which works. Henceforth assume $f$ is nonconstant and $f$ is injective at $0$. Note that if $y=\frac{x}{f(x)-x}$, then $yf(x)=x(y+1)$. If $f(x)\ne x$ for some $x$, $P\left(x,\frac{x}{f(x)-x}\right): \left(\frac{x}{f(x)-x}+1\right)f\left(\frac{xf(x)}{f(x)-x}\right) = \frac{x}{f(x)-x}\cdot f\left(\frac{xf(x)}{f(x)-x}\right)$ Since $\frac{x}{f(x)-x}\ne \frac{x}{f(x)-x} + 1$, we have \[f\left(\frac{xf(x)}{f(x)-x} \right) = 0,\]so $f(x)=0$, which implies $x=0$. Thus, $f(x)=x$ for all $x\ne 0$, and clearly $f(0)=0$. So $\boxed{f(x)=x}$, which works.