Let $P(x,y)$ be the assertion We see that $P(0,0): f(f(0))=0$ Let us have $P(0,x):f(f(x))=x$ So the function is injective. Let us write $f(0)=a$ so $f(a)=0$ and $P(a,a): f(2a^3)=a$ or $2a^3=0$ giving $f(0)=0$ Now we take $P(x,0):f(2x^3)=2x^2f(x)$ So the equation becomes $f(2x^3+f(y))=y+f(2x^3)$ and taking $P(x,f(y)): f(2x^3+f(f(y)))=f(y)+f(2x^3)$ $f(2x^3+y)=f(y)+f(2x^3)$ Since every real number can be written as $2x^3$, the function is of Cauchy form and from $f(f(x))=x$ we get $c^2=1$ and therefore $f(x)=x$ and $f(x)=-x$. Which are indeed solutions

Com10atorics wrote: $f(2x^3+y)=f(y)+f(2x^3)$ Since every real number can be written as $2x^3$, the function is of Cauchy form and from $f(f(x))=x$ we get $c^2=1$ and therefore $f(x)=x$ and $f(x)=-x$. Which are indeed solutions It's true that $f(x+y)=f(x)+f(y)$, however, it doesn't imply $f$ is linear. You'll need to prove other conditions, CMIIW.

any ideas?

I give the last step. Since $f(x+y)=f(x)+f(y)$ we have that $f(2x)=2f(x)$ so $f(x^3)=x^2f(x)$ and $f(qx)=qf(x),q\in \mathbb{Q}$ $x^2f(x)+3f(x^2)+3f(x)+f(1)=f((x+1)^3)=(x+1)^2f(x+1)=(x+1)^2(f(x)+f(1)\,\,\,(1)$ $fx^2f(x)-3f(x^2)+3f(x)-f(1)=f((x-1)^3)=(x-1)^2(f(x)-f(1))\,\,\,\,(2)$ $(1)+(2)\Rightarrow ...\Rightarrow f(x)=cx$ etc

$f(2x^3+f(y))=y+2x^2f(x)$ it is obvious that $f$ is bijective $P(0,0)$ $ff(0)=0$ $P(x,0)$ $ff(x)=x$ $P(\sqrt{\frac{1}{2}},0)$ $f(0)=0$ $P(x,f(-2x^3))$ $f(-2x^3)=-2x^2f(x)$ $P(x,f(y-2x^3))$ $f(-2x^2f(x))=-2x^3$ $P(f(x),f(-2f(x)^3))$ $f(-2f(x)^3)+2x^3=0$ $f(-2f(x)^3)=f(-2x^2f(x))$ $f(x)=+-x$