Ukrainian mathematical olympiad 2018 10.4 and 11.3. See also.

Thanks! I didn't know that this problem was also given on another competition.

@Pinko I do not think 2) in your solution is correct, you implicitly assume $f(0) = 0$ which is not given or proven before. Moreover, later on you prove that if $z$ is such that $f(z) \neq 0$ then $f(2z) = 0$, so overall you have "for each $x$ we have $f(x) = 0$ or $f(2x) = 0$" but this does not imply that $f(2x) = 0$ for all $x$ - e.g. the function $f(1) = 1$, $f(4) = 100$, $f(x) = 0$ otherwise satisfies this (and there are many more examples)

Here's a full solution, fixing Pinko's gaps. As $f\equiv 0$ is a solution, assume, for contradiction, that there is some $b$ with $f(b) > 0$. We firstly determine $f(0) = a$. Denote the assertion by $P(x,y)$. $P(x,0)$ gives $f(f(x) + a) = 0$. Hence $P(x+f(a), 0)$ gives $0 = f(f(x+f(a)) + a) = f(a)$. Then $P(f(x), a)$ leads to $f(f(f(x))) = 0$ (using $f(f(x) + a) = 0$). In particular $f(f(f(0))) = 0$ and since $f(f(0)) = 0$, we deduce $f(0) = 0$ and $b\neq 0$. Moving on, $P(x,0)$ yields $f(f(x)) = 0$ and hence $P(x, f(y))$ gives $xf(y)f(x+f(y)) = f(f(x) + f(f(y))) = f(f(x)) = 0$. In other words, if $f(y) > 0$, then $f(x+f(y)) = 0$ for all $x$ and hence $f(f(x) + f(y)) = 0$ for all $x$. With the latter, $P(x,b)$ implies $0 = f(f(x) + f(b)) = xbf(x+b)$ and so $f(x+b) = 0$ for all $x>0$, i.e. $f(x) = 0$ for all $x>b$. In particular, $b$ must be unique (as if $b_1>b_2$ are such that $f(b_1) > 0$, $f(b_2) > 0$, then $x=b_1$ gives a contradiction). But then $f(\frac{b}{2}) = 0$ and $P(\frac{b}{2}, \frac{b}{2})$ yields $0 = f(2f(\frac{b}{2})) = \frac{b^2}{4}f(b)$ which with $b>0$ implies $f(b) = 0$, contradiction! Therefore $f \equiv 0$ is the only solution.

Under revision.

jasperE3 wrote: Here's a quick proof. $P(1,-1)\Rightarrow f(f(1)+f(-1))+f(0)=0$ and since $f(x)\ge0$ equality holds and $f(0)=0$. Ummm Pinko wrote: Find all functions $f :[0, +\infty) \rightarrow [0, +\infty)$ for which $f(f(x)+f(y)) = xy f (x+y)$ for every two non-negative real numbers $x$ and $y$.