Calculus (3rd Edition)

$$h_z(3,0) =3 .$$
Since $h(x,z))=e^{xz-x^2z^3}$, then by the chain rule, we have $$h_z=e^{xz-x^2z^3}(x-3x^2z^2)=(x-3x^2z^2)e^{xz-x^2z^3}.$$ Hence, we get $$h_z(3,0)=(3-0)e^{0}=3 .$$