58th Ukrainian National Mathematical Olympiad

For $y=0$ from the condition we have, that $f(f(x)+f(0))=0$ for any $x$, and there is such $a$ that $f(a)=0$. Thus, for $x=a$, $f(f(0))=0$. Then, for $x=y=0$ it follows, that $f(2f(0))=0$, and for $x=y=f(0)$ we have, that $$f(f(f(0))+f(f(0)))=f^2(0)f(2f(0))=0,$$ i.e. $f(0)=0$ and $f(f(x))=0$ for any $x$.

Let $y=f(y)$ in the condition, then $$xf(y)f(x+f(y))=f(f(x)+f(f(y)))=f(f(x))=0,$$ so $xf(y)f(x+f(y))=0$ for arbitrary $x$ and $y$.

Let $f(y_0)\ne 0$ for some $y_0>0$. Let $y=y_0-x$, then $$f(f(x)+f(y_0-x))=x(y_0-x)f(y_0).$$ Expression in right part takes arbitrary value in segment $[0;\frac{1}{4}{y}_0^{2}f(y_0)]$, so range of function $f$ includes this segment. Thus, there is some $x_0$ such, that $f(x_0)=\min \{\frac{1}{2}{y}_0,\frac{1}{8}{y}_0^{2}f(y_0)\}$. From equality $xf(y)f(x+f(y))=0$ for $y=x_0$ and $x=y_0-f(x_0)>0$ we have, that $f(x_0)f(y_0)=0$, that is impossible. This contradiction proves, that $f(x)=0$ for arbitrary $x\in [0,+\infty )$.