1)$f(0)=0, x=0 \Longrightarrow f(y)=0$ 2)$f(0)\neq 0\Longrightarrow y=0 \Longrightarrow$ $f$ is injective. $x=2\Longrightarrow f(2)=1$ absolutely $f(x)=1$ is not solution.so there exist $x_0$ such that $f(x_0)\neq 1$ (it is obvious $x_0\neq 2$) $y=\frac{2f(x_0)-x_0}{f(x_0)-1} \Longrightarrow (x_0-2)f(\frac{2f(x_0)-x_0}{f(x_0)-1})=0\Longrightarrow f(\frac{2f(x_0)-x_0}{f(x_0)-1})=0$ $x=\frac{2f(x_0)-x_0}{f(x_0)-1} \Longrightarrow f(y)(\frac{f(x_0)-x_0+1}{f(x_0)-1})=0$ 1)$f(x_0)-x_0+1\neq 0 \Longrightarrow f(y)=0$ 2)$f(x_0)=x_0-1 \Longrightarrow \frac{2f(x_0)-x_0}{f(x_0)-1}=1 \Longrightarrow f(1)=0$ $y=1\Longrightarrow f(x)=x-1$ ======================================================== solutions: $1)f(x)=0$ $2)f(x)=x-1$

Here is my solution :

$\spadesuit \color{green}{\textit{\textbf{ANS:}}}$ $f(x)=0 \quad f(x)=x-1 \quad \forall x\in \mathbb{R}.$ $\clubsuit \color{red}{\textit{\textbf{Proof:}}}$ It is easy to see that these are indeed solutions to the given FE. Let $P(x,y)$ denote the given assertion, if $f(0)=0$, then \[P(0,x): \boxed{f(x)=0 \quad \forall x\in \mathbb{R}}.\]If $f(0)\ne 0$, then \[P(x,0): (x-2)f(0)+f(2f(x))=f(x)\]which gives injectivity for $f$. Then, \[P(2,0):f(2f(2))=f(2) \implies f(2)=1.\]For all $x\ne 2$, we have \[P\left(x,\frac{2f(x)-x}{f(x)-1}\right): f\left(\frac{2f(x)-x}{f(x)-1}\right)=0.\]Now, as $f(3)\ne 1$, \[P\left(3,\frac{3}{1-f(3)}\right): f\left(\frac{3}{1-f(3)}+2f(3)\right)=0\]\[P\left(3,\frac{3-2f(3)}{1-f(3)}\right): f\left(\frac{3-2f(3)}{1-f(3)}\right)=0\]by injectivity, we have $\frac{3}{1-f(3)}+2f(3)=\frac{3-2f(3)}{1-f(3)} \implies f(3)=0 \quad \textrm{or} \quad 2.$ If $f(3)=0$, then as $f\left(\frac{2f(x)-x}{f(x)-1}\right)=0$, \[2f(x)-x=3f(x)-3 \implies f(x)=3-x\]but by checking, this is not a solution. If $f(3)=2$, then $f(1)=0$ and by the same argument, we have \[2f(x)-x=f(x)-1 \implies f(x)=x-1\]which is a solution indeed. $\quad \blacksquare$