Ukrajina 2008

Let's make substitution $x=-2-t$ where $t$ is an arbitrary real number. We find that $f(f(y)-t)+f(f(y)+2+t)=-yf(y)(t+1)$. We can see that the left side of the equation has not changed while a minus sign appeared on its right side. Thus for all $y$ and $t$ the following equation should be true: $yf(y)(t+1)=0$, which implies that $f(y)=0$ for all $y\ne 0$. Let's make another substitution $x=-2$, $y=1$. We find: $f(0)+f(2)=0$. As $f(2)=0$, $f(0)=0$. Evidently, the identically zero function fulfills the condition.