Answer
$$ h_{vv}(u,v)= \frac{32u}{(u+4v)^3}$$
Work Step by Step
Since $ h(u,v)=\frac{u}{u+4v}=u(u+4v)^{-1}$, then by using the chain rule, we have
$$ h_v(u,v)=\frac{\partial h}{\partial v}=- u(u+4v)^{-2}(4)=-\frac{4u}{(u+4v)^2}.$$
Hence, again by using the chain rule, we get
$$ h_{vv}(u,v)=\frac{\partial h_v}{\partial v}=8 u(u+4v)^{-3}(4)=\frac{32u}{(u+4v)^3}$$
