egxa wrote:

Find all $f:\mathbb{R}\to\mathbb{R}$ functions such that
$f(x+y)^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)$ for all real numbers $x,y$

Let $P(x,y)$ be the assertion $f(x+y)^3=(x+2y)f\left(x^2\right)+f(f(y))\left(x^2+3xy+y^2\right)$ .
First, we show that $f(x)^3=xf\left(x^2\right)=x^2f(f(x))$ , which in particular implies $f(f(1))=f(1)$ :
$P(0,0)\Rightarrow f(0)=0$
$P(x,0)\Rightarrow f(x)^3=xf\left(x^2\right)$
$P(0,x)\Rightarrow f(x)^3=x^2f(f(x))$

Because of this, we can rewrite $P(x,y)$ as:
$f(x+y)^3=f(x)^3+f(y)^3+2yf\left(x^2\right)+(x^2+3xy)f(f(y)).$ Swapping $x,y$ and comparing, this becomes:
$2yf\left(x^2\right)+(x^2+3xy)f(f(y))=2xf\left(y^2\right)+(y^2+3xy)f(f(x)),$ and setting $y=1$ and simplifying:
$2x^2f\left(x^2\right)+(x^4+3x^3)f(f(1))=2x^3f(1)+(3x^3+x^2)f(f(x))$ $2xf(x)^3+(x^4+3x^3)f(1)=2x^3f(1)+(3x+1)f(x)^3$ which gives $f(x)^3=x^3f(1)$ for all $x\ne-1$ .

Then, $f(1)^3=f(1)$ , so $f(1)\in\{-1,0,1\}$ .
If $f(1)=-1$ then $f(-1)^3=-f(1)=1$ (using $f(x)^3=xf\left(x^2\right)$ ) so $f(-1)=1$ , giving the solution $\boxed{f(x)=-x}$ .
If $f(1)=0$ then $f(-1)^3=-f(1)=0$ so $f(-1)=0$ , giving the solution $\boxed{f(x)=0}$ .
If $f(1)=1$ then $f(-1)^3=-f(1)=-1$ so $f(-1)=-1$ , giving the solution $\boxed{f(x)=x}$ .
Easy to verify that these all satisfy the equation.