Let $P(x, y)$ be the proposition that $f(f(x)+f(y))=xyf(x+y)$. Let $f(0) = a$. \begin{align*} P(0, y) :\ & f(a+f(y))=0\\ P(0, a+f(y)) :\ & f(a+f(a+f(y))) = 0\\ &\ f(a)= 0\\ P(a, f(y)) :\ & f^3(y) = af(y)f(a+f(y)) = 0\\ \end{align*}Substituting $y=0$ into the last statement gives $f^3(0) = 0$, but $f(a) = f^2(0) = 0$, hence $f(0)=0$. \begin{align*} P(0,y) :\ & f(f(y))=0\\ P(f(x),y) :\ & yf(x)f(f(x)+y) = f(f(f(x))+f(y)) = f(f(y)) = 0\\ \end{align*}In the last statement, if $y > 0$, $$0 = f(x)f(f(x)+y)$$In other words, if $f(x) > 0$, $f(f(x)+y) = 0$ for all nonnegative $y$. Now consider some $k$ such that $f(k) > 0$. \begin{align*} P(k, y): f(f(k)+f(y)) &= kyf(k+y)\\ 0 &= kyf(k+y)\\ \end{align*}Which means for any $j > k$, $f(j) = 0$. Hence there exists at most $1$ such $k$. $$P(k, k): f(2f(k)) = k^2f(2k) = 0$$Hence $f(k) \ne \frac k 2$. Also, since $f(f(k)) = 0$, $f(k) \ne k$. Hence either $f(x) = 0$, or there exists some $k \ne 0$ where $f(k) = c \ne 0, k, \frac k 2$, and $f(x) = 0$ otherwise. (Edit Oopsie, see #6 below) It can be checked that these functions work.

joyce_tan wrote: Substituting $y=0$ into the last statement gives $f^3(0) = 0$, but $f(a) = f^2(0) = 0$, hence $f(0)=0$. srr if its a dumb question but how do we get that $f(0) = 0$ shouldn`t we prove function is injective first?

@Above $0=f^3(0)=f(f^2(0))=f(0)$

Mindstormer wrote: Find all functions $f:[0,+\infty) \mapsto [0,+\infty)$, which for all nonnegative $x,y$ satisfy $$f(f(x)+f(y))=xyf(x+y)$$ Proposed by Igor Voronovich I think, the following problem will be more complicated: find all functions $f\colon (0,+\infty)\to(0,+\infty)$ such that $$f(f(x)+f(y))=xyf(x+y).$$Here we get non-trivial solution $f(x)=\frac 1 x$. Is it true that there are no more solutions?

joyce_tan wrote: Let $P(x, y)$ be the proposition that $f(f(x)+f(y))=xyf(x+y)$. Let $f(0) = a$. \begin{align*} P(0, y) :\ & f(a+f(y))=0\\ P(0, a+f(y)) :\ & f(a+f(a+f(y))) = 0\\ &\ f(a)= 0\\ P(a, f(y)) :\ & f^3(y) = af(y)f(a+f(y)) = 0\\ \end{align*}Substituting $y=0$ into the last statement gives $f^3(0) = 0$, but $f(a) = f^2(0) = 0$, hence $f(0)=0$. \begin{align*} P(0,y) :\ & f(f(y))=0\\ P(f(x),y) :\ & yf(x)f(f(x)+y) = f(f(f(x))+f(y)) = f(f(y)) = 0\\ \end{align*}In the last statement, if $y > 0$, $$0 = f(x)f(f(x)+y)$$In other words, if $f(x) > 0$, $f(f(x)+y) = 0$ for all nonnegative $y$. Now consider some $k$ such that $f(k) > 0$. \begin{align*} P(k, y): f(f(k)+f(y)) &= kyf(k+y)\\ 0 &= kyf(k+y)\\ \end{align*}Which means for any $j > k$, $f(j) = 0$. Hence there exists at most $1$ such $k$. $$P(k, k): f(2f(k)) = k^2f(2k) = 0$$Hence $f(k) \ne \frac k 2$. Also, since $f(f(k)) = 0$, $f(k) \ne k$. Hence either $f(x) = 0$, or there exists some $k \ne 0$ where $f(k) = c \ne 0, k, \frac k 2$, and $f(x) = 0$ otherwise. It can be checked that these functions work. By $P(\dfrac{k}{2}, \dfrac{k}{2})$ we will get that $0 = \dfrac{k^2}{4}f(k)$ so $f(k) = 0$ and this is contradiction with assumption that exists a positive value of $f$, so its constantly zero.

@joice_tan, after your edit its still incorrect: there is no non-zero solution, check my comment properly

Yeah, I wanted to leave my original wording there, only in red, just so it's not that confusing when you quote it. I realize that $f(x) = 0$ is the only solution.

I don't think that problem must be such easy.Igor Voronovich is master in FE and Mindstormer titled this problem as R+ so maybe this question was proposed as R+ in contest.Am i true about it?

@above the problem is exactly what I wrote; my mistake though, changed the title from R+ to R_>=0