Taking $x=0$ we find $f(f(y)) = y$, so $f$ is bijective. Let $x_0$ be such that $f(x_0)=0$. Taking $y=x_0$, we find $f(x^2) = -x^2 + x_0$, so $f(t) = -t+ x_0$ for any $t\ge 0$. Now fix any $y=y_0$ and take $x$ large enough, so that $x^2+f(y_0)>0$. We get $f(x^2+f(y_0)) = -x^2-f(y_0) +x_0 = y_0-x^2$. So, $f(y_0) = -y_0+x_0$ for any $y_0$. Thus, the only answers are $f(x) = -x+c$ where $c$ is an arbitrary constant independent of $x$.