How find the function $f(f(x)+xy)=f(x)+xf(y),\forall x,y\in R$ Find all function $f:R\to R$ and such
$$f(f(x)+xy)=f(x)+xf(y),\forall x,y\in R$$
Let $x=y=0$,then $$f(f(0))=f(0)$$

Let denote that $f^0(x) = x$ and that for positive integer $n$, $f^n(x)=f(f^{n-1}(x))$.
It is easily proven by direct calculation that for every positive integer $n$ and every real $x$, $f^n(x)=f(x)+xf(0)$.
This fact implies that $f(f(x)-x)=f(x)-x +f(x)f(0)-xf(0)$.
But substituting $y = -1$, for every real $x$ provides $f(f(x)-x)=f(x)+xf(-1)$.
Thus we get that  $[1+f(0)+f(-1)]x=f(x)f(0)$ which shows that $f(0)=0$.
Using $f(0)=0$, we can see that for every positive integer $n$ and every real $x$, $f^n(x)=f(x)$.
So $f(f(x)-x)=f(x) - x$. 
Since $f(f(x)-x)=f(x)+xf(-1)$, it leads to $f(-1)=-1$.
Now we have that for every real $x$, $f(f(x)-x)=f(x)- x$, and this implies that $f(f^2(x)-f(x))=f^2(x)-f(x)$ equivalent to $f(x)-x=f(0)=0$.
Therefore $f(x)=x$.
Notice that the case $f(x)=C$ for $C$ a real constant, we get that $C=C+Cx$. Thus $C = 0$ and $f(x)=0$.
