Follow up to previous functional equation question: $f\big(x^2+f(y)\big)=(x-y)^2f(x+y)$ 
Find all $f:\mathbb{R} \to \mathbb{R}$ s.t. $$f\big(x^2+f(y)\big)=(x-y)^2f(x+y)$$

Putting $x=y$ yields $$f\big(x^2+f(x)\big)=0\text.\tag1\label1$$
Putting $y=-x$ yields $$f\big(x^2+f(-x)\big)=(2x)^2f(0)\text.\tag2\label2$$
By \eqref{1}, $(2x)^2f(0)=0$ ($\forall x \in \mathbb{R}$)
$$\implies \; f(0)=0$$
Hence putting $y=0$ we have $f\big(x^2+f(0)\big)=f(x)x^2$ which from above is $f\big(x^2\big)=f(x)x^2$.
Using this last equation we have $f(a)=af\left(a^{\frac12}\right)=a^{1+\frac12}f\left(a^{\frac14}\right)=a^{1+\frac12+\frac14}f\left(a^{\frac18}\right)= \cdots =a^2f\left(a^0\right)=a^2f(1)$.
So $f(x)=kx^2$ for some real constant $k$.
Conversely we check and see that it works only for $k=0$ (I think).
My question essentially is - in the sort of limit part is that rigorous (I'd love to avoid explicit calculus) and is my final deduction correct?
Thanks for any help.
 A: There's a way to prove $f(x)=0$ or $f(x)=-x^2$ without using continuity.
Put $-x$ in the place of $x$ and you'll get:
$$f\big((-x)^2+f(y)\big)=(-x-y)^2f(-x+y)$$
$$\therefore(y+x)^2f(y-x)=(y-x)^2f(y+x)$$
Now put $\frac{y+x}{2}$ and $\frac{y-x}{2}$ in the place of $y$ and $x$, respectively, and you'll have:
$$y^2f(x)=x^2f(y)$$
$$\therefore f(x)=x^2f(1)$$
Put $k=f(1)$ and check if the answer works:
$$k\big(x^2+ky^2\big)^2=(x-y)^2k(x+y)^2=k\big(x^2-y^2\big)^2$$
$$\therefore k(1+k)y^2\big(2x^2+(k-1)y^2\big)=0$$
The above equation holds for $x=1$ and $y=1$, so the only valid values for $k$ are $0$ and $-1$.
A: You obtained tw0 relations: $(1)$ $f(x^{2}+f(x))=0$ and $(2)$ $f(x^{2})=x^{2}f(x)$ , but for complete of solution you need another relation, the functional equation ( or recently relation) show that the function $f$ is even. Also, It's obvious that there is a kind of symmetric relation in functional equation, so we have
 $$(3)\ \ f(x^{2}+f(y))=f(y^{2}+f(x))$$
The relations $(2)$ and $(3)$ imply that 
$$(4)\ \ f(f(x))=x^{2}f(x)$$
We know that $f(0)=0$ and $f$ is an even function, so if the function $f$ is a nonzero function, then relation $(4)$ implies that the point $x=0$ is only root of $f$, therefore relation $(1)$ implies that $f(x)=-x^{2}$ and the proof is done.    
