Plug in $y=-x$, and let $c=f(0)$. We have $f(x^2) = c^2/2$. Plugging in $y=0$, we have $f(x)^2 = c^2/2 + c$. Plugging in $x=0$ in this gives $c=0$ or $c=2$. Now if $c=0$, $f \equiv 0$ is the only solution. Else we have $f(x) = \pm 2$ for all $x$. This means $f(x^2) + f(y^2) = 4$, always, so for all $x$ being perfect squares, $f(x)=2$. So the solution in this case becomes $f(x) = 2$ for all perfect square $x$ and $\pm 2$ for all other $x$. It is easy to check that the solutions mentioned above work.

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