jmc

Setting $y=0,$ we get $$f(x^2)=f(x{)}^2-2xf(0).$$Let $c=f(0),$ so $f(x^2)=f(x{)}^2-2cx.$ In particular, for $x=0,$ $c=c^2,$ so $c=0$ or $c=1.$

Setting $x=0,$ we get $$f(y^2)=c^2+y^2.$$In other words, $f(x^2)=x^2+c^2$ for all $x.$ But $f(x^2)=f(x{)}^2-2cx,$ so $$f(x{)}^2-2cx=x^2+c^2.$$Hence, $$f(x{)}^2=x^2+2cx+c^2=(x+c{)}^2.(\ast )$$Setting $y=x,$ we get $$c=f(x{)}^2-2xf(x)+x^2,$$or $$f(x{)}^2=-x^2+2xf(x)+c.$$From $(\ast ),$ $f(x{)}^2=x^2+2cx+c^2,$ so $-x^2+2xf(x)+c=x^2+2cx+c^2.$ Hence, $$2xf(x)=2x^2+2cx=2x(x+c).$$So for $x\ne 0,$ $$f(x)=x+c.$$We can then extend this to say $f(x)=x+c$ for all $x.$

Since $c$ must be 0 or 1, the only possible solutions are $f(x)=x$ and $f(x)=x+1.$ We can check that both functions work.

Thus, $n=2$ and $s=1+2=3,$ so $n\times s=\boxed{6}.$