If we put $x=1$ we get: \[ f(y+1)= f(y)+2 \;\;\;(1)\] Let $a=f(0)$ and $b=f(1)$, then $b=a+2$. If we put $y=1$ we get: \[ xb+f(xb)-xf(b)-f(x)= 2x+b-f(x+1) \;\;\; (2)\] If we put (1) in (2) we get: \[ xa+f(xb)=xf(b)+a\] So if $b\ne 0$ we have $f(x)= x\cdot {f(b)-a\over b}+a$ If we put here $x=1$ we get $f(b)=3b-2$, so $f(x)=2x+a$. If $b=0$ then $a=-2$. Then from $(1)$ we get $f(2)=2$: Put $y=2$ in original equation and we get: $f(x+2)= 2x+2$, so $f(x)=2x-2$. So solution is $\boxed{f(x)=2x+a}$ for every $a\in \mathbb{R}$. (We should check this.)

The answers is $f(x)=2x+f(0)$

Let $P(x,y)$ $ \Rightarrow $ \[ xf(y)+f(xf(y))-xf(f(y))-f(xy)=2x+f(y)-f(x+y). \] Then we have $P(1,x)$ $ \Rightarrow $ \[ f(x+1)=f(x)+2 \] for all $x$ and $P(x,1)$ $ \Rightarrow $ \[ f(xf(1))=x(f(f(1))-f(1)+2)+f(0) . \] If $f(1)\not= 0$ then we get that $f$ is linear and easily get that $f(x)=2x+d$, there $d=f(0)$ is a real constant. Otherwise, $f(1)=0$ and $P(x,2)$ $ \Rightarrow $ \[ f(x)=2x-2 \] for all real $x$.

mathuz wrote: If $f(1)\not= 0$ then we get that $f$ is linear and easily get that $f(x)=2x+d$, there $d=f(0)$ is a real constant. Could please explain how did you get $f(x)=2x+d$

If I sub f(x)=2x+f(0) in, I get f(0)=0.

sub: xf(y)=2xy+xf(0) f(xf(y))=4xy+2xf(0)+f(0) xf(f(y))=4xy+2xf(0)+f(0) f(xy)=2xy+f(0) RHS=0, obvious Hence (x-1)f(0)=0? hence f(0)=0? Please enlighten me.

$$f(x)=2x+a$$