Strategies to find the set of functions $f:\mathbb R\to\mathbb R$ satisfing the functional equation: $f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy))$ My question is as follows: What methods can be used to find the set of functions $f:\mathbb{R}\to\mathbb{R}$ satisfying a certain functional equation. An example of a case where this applies is the following:

Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following equation:
$$f\big(x^3\big)+f\big(y^3\big)=(x+y)\Big(f\big(x^2\big)+f\big(y^2\big)-f(xy)\Big):\forall x, y\in\mathbb R$$

I'm curious as to whether there are general methods (or strategies) for solving this type of question, or whether questions like these should just be handled on a case-by-case basis.
Thanks in advance.
 A: 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.
A: 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}
A: 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 $1$.
