find all $f:\mathbb{R} \rightarrow \mathbb{R} $ that satisfies,
$\displaystyle xf(x+xy)=xf(x)+f(x^2)f(y)$
for any real number $x,y$
It's clear that $f(0)=0$ and $f(y+1)=f(1)+f(1)f(y)\rightarrow f(1)=0$ or $f(-1)=-1$
if $f(1)=0$ it's not hard to show that $f(x)=0$ is the solution.
if $f(1)\not =0$ then $f(-1)=-1$ we plug $y=-1$ into the equation, that is, $f(x^2)=xf(x)..(i)$
Thus, for any $x\not = 0$,
$\displaystyle xf(x+xy)=xf(x)+f(x^2)f(y)=xf(x)+xf(x)f(y)\rightarrow f(x+xy)=f(x)+f(x)f(y)..(j)$
since, $(i)$, we have that $-xf(-x)=f(x^2)=xf(x)$ so $f(x)=-f(-x)$ for any $x$, that is, $f(1)=1$
We also clear that $f(x+1)=1+f(x)$ and $f(2x)=2f(x)$. consider from the equation $(j)$ that
$\displaystyle \Big(1+f(x)\Big)^2=1+2f(x)+2f(x)\Big(\frac{f(x)}{2}\Big)=1+f(2x+2x\cdot \Big(\dfrac{x}{2}\Big))=1+f(x^2+2x)$
$\displaystyle =f(1+x^2+2x)=f\Big((x+1)^2\Big)=(x+1)f(x+1)=(x+1)\Big(1+f(x)\Big)$
It's not hard to show that $f(x)=x$ is the another solution.