Strategies to find the set of functions $f:\mathbb R\to\mathbb R$ satisfing the functional equation: $f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy))$

Here is a typical start. Put $x=y=0$. The right side is $0$, so $f(0)=0$.

Now set $y=0$, and let $x$ roam freely. Since $f(0)=0$, we get $f(x^3)=xf(x^2)$.

Set $y=-x$. We get $f(x^3)+f(-x^3)=0$. Since everything is a cube, we have $f(-u)=-f(u)$ for all $u$.

Now explore $x=y$. Can we learn anything from setting $x$ and/or $y$ qual to $1$?

A little playing has gotten us a lot of information, enough that we should be able to complete things.

Using the equation, we can see that $f(0) = 0$ is necessary. Let $y = -x$, then \begin{gather} f(x^3) + f(-x^3) = 0 \end{gather} This implies that $f$ must be an odd function. Also letting $y = x$ we see that \begin{equation} f(x^3) = x f(x^2) \end{equation} Plugging this relationship into the LHS gives \begin{align} xf(x^2) + yf(y^2) & = (x + y)(f(x^2) + f(y^2) - f(xy)) \\ 0 & = x f(y^2) + y f(x^2) - (x+y)f(xy) \tag{*} \end{align} If we let $y \to -y$ then by the symmetry of the function we have \begin{align} 0 = xf(y^2) - yf(x^2) + (x-y)f(xy) \tag{**} \end{align} Adding $(^*)$ and $(^{**})$ we see that \begin{equation} x f(y^2) = y f(xy) \end{equation} If we let $y = 1$ we obtain \begin{equation} f(x) = x f(1) \end{equation}

EDIT: At this point we should check whether this function indeed satisfies the above relation. Letting $f(1) = c$, and plugging in we have \begin{equation} c(x^3 + y^3) = c(x+y)(x^2 + y^2 - xy) = c(x^3 + xy^2 -x^2 y + x^2 y + y^3 - xy^2) = c(x^3 + y^3) \end{equation}

If it is of any help the only continuous solutions are of the form $f(x)=xf(1)$. Indeed we can use the property $f(x^3)=xf(x^2)$ to show $$f(x)=x^\frac{1}{3}f(x^\frac{2}{3})=x^\frac{1}{3}x^\frac{2}{9}f(x^\frac{4}{9})=\cdots=x^{\frac13\sum_i(\frac23)^i}f(1)=xf(1).$$ In fact we only need continuity at $1$.