Let $X(u, v)$ be a parameterization of a surface $M$ so that the first fundamental form is $I = h^2(du^2 + dv^2)$. Show that the Gauss curvature
$K = -\frac{1}{h^2}(\frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial v^2})\log h$
I don't know where to really start. I took the derivative of $I$ and I've been looking at the Gauss-Bonnet Theorem and other theorems in my notes but I've been stuck on this for a while now. Personally, this is a very difficult section for me in the class and no tutoring is offered for this class at my school. Can someone please help?
Suppose $\dim{M}=2$ and the metric $ds^2= E(u, v)du^2+G(u, v)dv^2$. Then the Gauss curvature $K$ is given by
$$K=−\dfrac{1}{2\sqrt{EG}} \left(\left(\dfrac{G_u}{\sqrt{EG}}\right)_u+\left(\dfrac{E_v}{\sqrt{EG}}\right)_v \right)$$
Then $E(u, v)=h^2(u, v)=G(u, v)$, imply
$$K=−\dfrac{1}{h^2} \left(\dfrac{\partial^2}{\partial u^2}+\dfrac{\partial^2}{\partial v^2} \right)\log{h}$$