I am asked to prove that for a surface in $\mathbb{R}^3$ with local coordinates in a chart, u,v, the coefficients of the second fundamental form can be calculated as follows:
eg.: (first entry in II) is the determinant of $$ x_{uu} \ \ x_u \ \ x_v\\ y_{uu} \ \ y_u \ \ y_v\\ z_{uu} \ \ z_u \ \ z_v\\ $$
divided by the root of the determinant of the first fundamental form. This is easy to do as far as I am allowed to assume that the norm of the normal vector given by $\partial_u \times \partial_v$ is constant. (ie $<L\partial_u,u>=-<D_{\partial_u}(N/norm(N),\partial_u>=-1/norm(N)<D_{\partial_u}N,\partial_u>$ Am I allowed to assume that? If not how can I prove this?
No, the norm of $\partial_u \times \partial_v$ is equal to $\sqrt{EG -F^2}$. It is not a constant on a general . Let $r(u,v)$ be a parametrization. $L = \langle r_{uu},n\rangle$ or equivalently $L=-\langle r_u,n_u\rangle$. As I already mentioned $$\|r_u \times r_v\| = \sqrt{EG -F^2}.$$ Then, $$L=\langle r_{uu},n\rangle=\langle r_{uu},\frac{r_u\times r_v}{\|r_u\times r_v\|}\rangle=\frac 1{\|r_u\times r_v\|} [r_{uu},r_u,r_v]= \frac 1{\sqrt{EG-F^2}} [r_{uu},r_u,r_v].$$
The last triple product is evaluated to the determinant you mentioned and gives you the result.