Let $P(x,y)$denote the assertion $f(x^3+xf(xy))=f(xy)+x^2f(x+y)$. We will prove $f(x)=0\forall x \in \mathbb{R}^{\ge 0}$ is the only function satisfying the condition.First clearly the function satisfys the condition we will prove no other function works. Let $x_0$ be a real root of $x^3+xf(1)-1$ clearly $x_0$ is positive now $P(x_0,\frac{1}{x_0} )$ tells us there is a positive $k$ so that $f(k)=0$. Let $x_1$ be a root of $x^3+xf(t)=k$ for some non-negative $t$ clearly$x_1$ is positive now $P(x_1,\frac{t}{x_1})$ implies $f(t)=0$. So $f(x)=0\forall x \in \mathbb{R}^{\ge 0}$. Remark: To make things simpler one can take $z=xy$ and try to write the equation in terms of $x,z$.

Taha1381 wrote: Let $x_1$ be a root of $x^3+xf(t)=k$ for some non-negative $t$ clearly$x_1$ is positive now $P(x_1,\frac{t}{x_1})$ implies $f(t)=0$. So $f(x)=0\forall x \in \mathbb{R}^{\ge 0}$. Can you explain this part please

MATH1945 wrote: Taha1381 wrote: Let $x_1$ be a root of $x^3+xf(t)=k$ for some non-negative $t$ clearly$x_1$ is positive now $P(x_1,\frac{t}{x_1})$ implies $f(t)=0$. So $f(x)=0\forall x \in \mathbb{R}^{\ge 0}$. Can you explain this part please $t$ was arbitary also is the sum of two non-negatove numberd equal zero they are both zero.If you want to know why such positive $x_1$ exists use the fact that $x^3+cx$ is surjective over non-real numbers.(to show this use continuty and the fact that the function is zero in zero and infinity at infinity.

WOW complicated.

Proposed by Mojtaba Zare.