It's already solved. The answer $f(x)=cx$

It's very easy problem.Here my solution: Let $ P(x,y)$ be the assertion of the function. $ P(0,0) $ _____ $ f(0)=0 $ $ P(x,x) $ ______ $ f(x^3)=xf(x^2) $ $ P(x,-x) $ _____ so $ f $ is odd function. So $ RHS=LHS= xf(x^2)+ yf(y^2) $ so we get new equation $ f(xy)(x+y)= yf(x^2)+ xf(y^2) $ Finally we comparing $ P(x,1) $ and $ P(-x,1) $ we get $ f(x)=cx $, where $ c=f(1) $. I’m sorry for my bad English.

1) $ x=y=0 :f(0)=0 $, 2) $x=x,y=0 :f(x^3)=xf(x^2)$, 3) $x=x,y=-x: f(-x)=-f(x) $ , 4) $x=x,y=y$ and $x=x,y=-y$ : $2f(x^3)=2xf(x^2)+2xf(y^2) - 2yf(xy)$, 5) from 2) and 4) : $xf(y^2)=yf(xy)$ , 6) 5) to $ x=1/y : f(y^2)=y^2f(1)$, 7) from 6) $x=y^2 : f(x)=xf(1)=xc$ 8) $f$ is odd function so :$f(-x)=-f(x)=-xf(1)=-xc$ so answer :$ f(x)=xc$ $(c=f(1))$

Let $P(x,y)$ be the assertion. $P(x,-x) \Rightarrow f(x^3)=-f(-x^3) \Rightarrow \boxed{f(-x)=-f(x)} \forall x \in \mathbb{R}$ $P(x,0) \Rightarrow \boxed{f(x^3)=xf(x^2)} \forall x \in \mathbb{R}$ Substituting $xf(x^2)=f(x^3)$ in the original function we get $xf(y^2)+yf(x^2)=(x+y)f(xy)$...(1) Putting $y=-y$ in (1) we get $xf(y^2)-yf(x^2)=(x-y)f(-xy)=(y-x)f(xy)$...(2) Adding (1) and (2) we get $xf(y^2)=yf(xy)$.Putting $y=1$ in this equation yeilds $\boxed{f(x)=cx} \forall x \in \mathbb{R}$ It is easy to see that this function indeed satisfies the relation.

What about f(x^3+y^3)=(x+y)(f(x^2)-f(xy)+f(y^2))

toto1234567890 wrote: What about f(x^3+y^3)=(x+y)(f(x^2)-f(xy)+f(y^2)) In $latex$:$f(x^3+y^3)=(x+y)(f(x^2)-f(xy)+f(y^2))$

Solution: Let $P(x,y)$ be the assertion. From $P(x,0)$ we have \[f(x^3)=xf(x^2) \] which is equivalent to \[f(m)=m^{\frac{1}{3}}f(m^{\frac{2}{3}})\] where $m\in\mathbb{R^{+}}$. If we iterate this relation indefinitely, we get \[f(m)=m^{S_1}f(m^{S_2}) \] where \[S_1=\frac{1}{3}+\frac{2}{9}+...=1 \] and \[S_2=\lim_{t \to \infty}{\frac{1}{(\frac{3}{2})^t}}=0\] so $f(m)=mf(1)$. Our solution for the positive reals $x$ is $f(x)=cx$, where $c$ is a real constant, which clearly fits the relation. Similarly, using $P(0,y)$, we can find the same solution for negative reals. Therefore, our only solution is $\boxed{f(x)=cx, c\in\mathbb{R}}$. $\blacksquare$

PlatinumFalcon wrote: Solution: Let $P(x,y)$ be the assertion. From $P(x,0)$ we have \[f(x^3)=xf(x^2) \] which is equivalent to \[f(m)=m^{\frac{1}{3}}f(m^{\frac{2}{3}})\] where $m\in\mathbb{R^{+}}$. If we iterate this relation indefinitely, we get \[f(m)=m^{S_1}f(m^{S_2}) \] where \[S_1=\frac{1}{3}+\frac{2}{9}+...=1 \] and \[S_2=\lim_{t \to \infty}{\frac{1}{(\frac{3}{2})^t}}=0\] so $f(m)=mf(1)$. Our solution for the positive reals $x$ is $f(x)=cx$, where $c$ is a real constant, which clearly fits the relation. Similarly, using $P(0,y)$, we can find the same solution for negative reals. Therefore, our only solution is $\boxed{f(x)=cx, c\in\mathbb{R}}$. $\blacksquare$ You can not get $f(m^{S_2})=f(1)$ unless you can show $f$ is continuous at $1$.

How could I fix that hole?

Unfortunately, you cannot. The entire proof relies on a faulty assumption of continuity. Although assuming continuity can tell you all the solutions of a functional equation, which might be helpful (since pretty much every olympiad FE has only continuous solutions). It's a very common mistake for young beginners like you, to fakesolve functional equations with these kind of assumptions. With practice and guidance, you will get better.

see also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2647093

Albanian old tst also see...... Probably the easiest...

Great Problem. Observe setting $x=y=0$ gives $f(0)=0$. Plugging $y=-x$ gives $f(x^3)+f(-x^3)=0$, or that $f$ is odd. Taking $y=0$, we get $f(x^3)=xf(x^2)$. Therefore, $$(x+y)(f(x^2)+f(y^2)-f(xy))=f(x^3)+f(y^3)=xf(x^2)+yf(y^2).$$Expanding both sides and simplifying, $$(x+y)f(xy)=yf(x^2)+xf(y^2).$$Taking $y=1$ gives $f(x^2)=(x+1)f(x)-xf(1)$, and letting $y=-1$ gives $f(x^2)=xf(1)-(x-1)f(-x)=xf(1)+(x-1)f(x)$. Setting the two expressions equal, we get $f(x)=xf(1)$, which implies the solution is $\boxed{f=cx}$.

see here: https://artofproblemsolving.com/community/c6h535689p3074555