jmc

Setting $x=y=0,$ we get $$f(0)=zf(0)$$for all $z,$ so $f(0)=0.$

Setting $y=0,$ we get $$f(x^2)=xf(x)$$for all $x.$

Setting $x=0,$ we get $$f(yf(z))=zf(y).$$In particular, for $y=1,$ $f(f(z))=zf(1).$

Since $f(x^2)=xf(x),$ $$f(f(x^2))=f(xf(x)).$$But $f(f(x^2))=x^{2}f(1)$ and $f(xf(x))=xf(x),$ so $$x^{2}f(1)=xf(x).$$Then for $x\ne 0,$ $f(x)=f(1)x.$ Since $f(0)=0,$ $$f(x)=f(1)x$$for all $x.$

Let $c=f(1),$ so $f(x)=cx.$ Substituting into the given equation, we get $$c{x}^2+c^{2}yz=c{x}^2+cyz.$$For this to hold for all $x,$ $y,$ and $z,$ we must have $c^2=c,$ so $c=0$ or $c=1.$

Thus, the solutions are $f(x)=0$ and $f(x)=x.$ This means $n=2$ and $s=0+5,$ so $n\times s=\boxed{10}.$