Are you sure the problem statement is correct? $P(0,0)\implies f(0)=0$ $P(x,0)\implies f(x)+f(-x)=2x\implies 2x=f(x)+f(-x)=f(-x)+f(x)=-2x$,a contradiction,so no function exists.

It should be $2xy$. I've edited.

The answer is $f(x)=x$ Any solution

FERZIO wrote: The answer is $f(x)=x$ Any solution Missing one solution Missing proof for the given solution ...

quangminhltv99 wrote: Find all $f:\mathbb{R}\to\mathbb{R}$ satisfying: $$f(xf(y)-y)+f(xy-x)+f(x+y)=2xy,\quad\forall x,y\in\mathbb{R}.$$ Let $P(x,y)$ be the assertion $f(xf(y)-y)+f(xy-x)+f(x+y)=2xy$ $P(0,0)$ $\implies$ $f(0)=0$ $P(x,0)$ $\implies$ $f(x)+f(-x)=0$ and so $f(x)$ is odd $P(-1,\frac 12)$ $\implies$ $f(u)=-1$ where $u=-f(\frac 12)-\frac 12$ $P(x,u)$ $\implies$ $f(x(u-1))=2ux$ This implies $u\ne 1$ and $f(x)=ax$ for some $a=\frac{2u}{u-1}$ Plugging this back in original equation, we get $a=\in\{-2,1\}$ and so : $\boxed{\text{S1 : }f(x)=x\quad\forall x}$ $\boxed{\text{S2 : }f(x)=-2x\quad\forall x}$

Let $P(x,y)$ the assertion of the given F.E. $P(0,x)$ $$f(-x)=-f(x) \implies f(0)=0$$$P(-1,x)$ $$f(x+f(x))=2x \implies f \; \text{surjective}$$Set $f(c)=-1$ and if $f(1)=-1$ then u get that $2=0$ so we have $c \ne 1$ and we do $P \left(\frac{x}{c-1},c \right)$ $$f(x)=\frac{2cx}{c-1}=c_1x$$Pluggin back in $P(-1,x)$ u get that $(c_1+2)(c_1-1)=0$ which means that $c_1=1$ or $c_1=-2$. Hence $f(x)=x$ or $f(x)=-2x$ thus we are done

Let $P(x,y)$ denote $f(xf(y)-y)+f(xy-x)+f(x+y)=2xy$. $P(0,0)$ implies $f(0)=0$. $P(x,0)$ implies $f(x)=-f(-x)$ and so $P(-1,1/2)$ implies $f(a)=-1$ for some $a$. If $a=1$ then $P(1,1)$ is contradiction, so $P(\tfrac{x}{a-1},a)$ implies $f$ is linear, which gives $f\equiv \text{Id}$ and $f(x)\equiv -2x$, as only working solutions.