Belorusija 2012

First, note that if $a+b+c=0$, then the function $f\equiv \lambda \ne 0$ satisfies the condition.

Now suppose that $a+b+c\ne 0$. In particular, at least one of $a,b,c$ is not zero. Say, $a\ne 0$. Let $(\ast )$ denote the given functional equation.

Set $x=y=0$ in $(\ast )$, then $af(f(z))=0$, or $f(f(z))=0$ for all $z\in \mathbb{R}$.

Now set $z=0$ in $(\ast )$, then $af(xy+f(0))=0$, or $f(xy+f(0))=0$. But $xy+f(0)$ takes all real values, hence $f$ is a zero constant. This contradiction does prove that $a+b+c=0$.