I'm trying to solve the functional equation $f(f(x)+xf(y))=xf(y+1)$.
Up to now I found that $f(f(0))=0$ when $x=0$ and that $f(y+1)=f(f(y)+f(1))$ by setting $x=1$. Also $f(x)=x$ is an apparent solution which might be the only one I think, but I'm not sure as I don't have any answers.
Could anybody help me with solving the problem?
Note that by setting $x=y=-1$ we see that $f(0)=-f(0)$ which implies that $f(0)=0$.
For $y=0$, we have that $f(f(x))=xf(1)$ so that $f(f(1))=f(1)$. We set $x=f(1)$ and get that $f(f(f(1)))=f(f(1))=f(1)=xf(1)=f(1)f(1)$ which implies that $f(1)=f(1)f(1)$ so that for non-zero $f(1)$, $f(1)=1$.
We now have that $f(f(x))=x$ so $f$ is an involution and is therefore injective. We set $x=1$ to get $f(y+1)=f(f(y)+f(1))$ which implies that $y+1=f(y)+f(1)$ because $f$ is injective. It follows from the fact that $f(1)=1$ that $f(y)=y$.
Finally, if $f(1)=0$ then $f(f(x))=0$. For $x=1$, $f(f(y))=f(y+1)$ so that $f(y+1)=0$ which implies that $f(x)=0$ which is another valid solution.