Plugging in $y$ as $0:$ \begin{equation}
f(x^2)=f(f(x))+f(0) \text{ } (1)
\end{equation}
Plugging in $x, y$ as $0:$ $$f(0)=f(f(0))+f(0)$$ or $$f(f(0))=0$$ Plugging in $x$ as $0:$ $$f(-y)+2yf(0)=f(f(0))+f(y),$$ but since $f(f(0))=0,$ \begin{equation}
f(-y)+2yf(0)=f(y) \text{ } (2)
\end{equation}
Plugging in $y^2$ instead of $y$ in the given equation: $$f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)$$ Replacing $y$ and $x$ : $$f(y^2-x^2)+2x^2f(y)=f(f(y))+f(x^2)$$ The difference would be:
\begin{equation}
f(x^2-y^2)-f(y^2-x^2)+2y^2f(x)-2x^2f(y)=f(f(x))-f(x^2)-f(f(y))-f(y^2) \text{ } (3)
\end{equation}
The right-hand side would be $f(0)-f(0)=0$ by $(1).$ Also, $$f(x^2-y^2)-f(y^2-x^2)=2(x^2-y^2)f(0)$$ by $(2)$ So, $(3)$ is reduced to: $$2(x^2-y^2)f(0)+2y^2f(x)-2x^2f(y)=0$$ Regrouping and dividing by 2: $$y^2(f(x)-f(0))=x^2(f(y)-f(0))$$ $$\frac{f(x)-f(0)}{x^2}=\frac{f(y)-f(0)}{y^2}$$ Because this holds for all x and y, $\frac{f(x)-f(0)}{x^2}$ is a constant. So, $f(x)=cx^2+f(0)$ .
This function must be even, so $f(y)-f(-y)=0$ .
So, along with $(2)$ , $2yf(0)=0$ for all $y$ , so $f(0)=0$ , and $f(x)=cx^2$ .
Plugging in $cx^2$ for $f(x)$ in the original equation, we get: $$c(x^4-2x^2y+y^2)+2cx^2y=c^3x^4+cy^2$$ $$c(x^4+y^2)=c(c^2x^4+y^2)$$ So, $c=0$ or $c^2=1.$ All of these solutions work, so the solutions are $f(x)=-x^2, 0, x^2$ .
-codemaster11