jmc

Let $y=\frac{x^2-f(x)}{2}.$ Then $$f\left(f(x)+\frac{x^2-f(x)}{2}\right)=f\left(x^2-\frac{x^2-f(x)}{2}\right)+4f(x)\cdot \frac{x^2-f(x)}{2}.$$Simplifying, we get $$f\left(\frac{x^2+f(x)}{2}\right)=f\left(\frac{x^2+f(x)}{2}\right)+2f(x)(x^2-f(x)),$$so $f(x)(x^2-f(x))=0.$ This tells us that for each individual value of $x,$ either $f(x)=0$ or $f(x)=x^2.$ (Note that we cannot conclude that the only solutions are $f(x)=0$ or $f(x)=x^2.$) Note that in either case, $f(0)=0.$

We can verify that the function $f(x)=x^2$ is a solution. Suppose there exists a nonzero value $a$ such that $f(a)\ne a^2.$ Then $f(a)=0.$ Setting $x=0$ in the given functional equation, we get $$f(y)=f(-y).$$In other words, $f$ is even.

Setting $x=a$ in the given functional equation, we get $$f(y)=f(a^2-y).$$Replacing $y$ with $-y,$ we get $f(-y)=f(a^2+y).$ Hence, $$f(y)=f(y+a^2)$$for all values of $y.$

Setting $y=a^2$ in the given functional equation, we get $$f(f(x)+a^2)=f(x^2-a^2)+4a^{2}f(x).$$We know $f(f(x)+a^2)=f(f(x))$ and $f(x^2-a^2)=f(x^2),$ so $$f(f(x))=f(x^2)+4a^{2}f(x).(\ast )$$Setting $y=0$ in the given functional equation, we get $$f(f(x))=f(x^2).$$Comparing this equation to $(\ast ),$ we see that $4a^{2}f(x)=0$ for all values of $x,$ which means $f(x)=0$ for all $x.$ We see that this function satisfies the given functional equation.

Thus, there are two functions that work, namely $f(x)=0$ and $f(x)=x^2.$ This means $n=2$ and $s=0+9=9,$ so $n\times s=\boxed{18}.$