Computing the Gaussian curvature of this surface $z=e^{(-1/2)(x^2+y^2)}$. Compute the Gaussian curvature of $z=e^{(-1/2)(x^2+y^2)}$. Sketch this surface and show where $K=0 $, $K>0$, and $K<0$.
So would the easiest way to do this question be to construct a parametrization $$\mathbf{x}(u,v)=(u, v, e^{-\frac{1}{2}(u^2+v^2)} )?$$
If so, I calculated the Normal to be $$\left( \frac{u}{u^2+v^2+e^{u^2+v^2}},  \frac{v}{u^2+v^2+e^{u^2+v^2}},  1 \right).$$ Is that correct? Thanks

If $g$ is the first fundamental form, and $h$ is the second fundamental form. Then we know
$$K = \frac{ \det h}{ \det g } $$
Since you have a graph $z=f(x,y)$, it's straight forward to compute the fundamental forms by definition. I'll leave $g$ to you, but $h$ is given by
$$ h = f_{xx} dx^2 + 2 f_{xy} dxdy + f_{yy} dy^2 $$
Hint: Consider $\gamma = ( x,y,f )$, by definition of the first fundamental form we have 
$$g_{ij} = ( \gamma_i , \gamma_j) $$
where the subscript denotes a derivative in $i$ and $( \cdot , \cdot)$ denotes the inner product. So we have
$$g = ( 1 + f_x^2 ) dx^2 + 2f_xf_y dxdy + ( 1 + f_y^2) dy^2 $$
