, 2008

Let's make substitution $x = -2 - t$ where $t$ is an arbitrary real number. We find that $f(f(y) - t) + f(f(y) + 2 + t) = -y f(y)(t + 1)$. 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: $y f(y)(t + 1) = 0$, which implies that $f(y) = 0$ for all $y \neq 0$.
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.