Derive the formula for Gaussian curvature of the surface $z(x,y)$ Now I am assuming that this problem is referring to a Monge patch i.e. $\sigma(x,y) = (x,y,z(x,y)).$
I know the Gaussian curvature of a Monge patch can be rewritten as 
$$k = \frac{z_{xx}z_{yy} - z_{xy}z_{yx}}{(1+z_x^2 + z_y^2)^2}$$ so this is the end goal.
 A: We have the following formula for the Gaussian curvature (see here):
$$K=\frac{LN-M^2}{EG-F^2}$$
where $E,G,F$ (respectively $L$, $M$, $N$) are the coefficients of the first fundamental form (respectively second fundamental form). 
Therefore, we just have to calculate the terms on the right hand side. Since
$\sigma_x=(1,0,z_x), \sigma_y=(0,1,z_y)$, we have 
$$E=\sigma_x\cdot\sigma_x=1+z_x^2, F=\sigma_x\cdot\sigma_y=z_xz_y, G=\sigma_y\cdot\sigma_y=1+z_y^2.\tag{1}$$
Moreover, the unit normal $\vec{n}$ is given by
$\vec{n}=\sigma_x\times\sigma_y=\displaystyle\frac{(-z_x,-z_y,1)}{\sqrt{1+z_x^2+z_y^2}}.$
Since $\sigma_{xx}=(0,0,z_{xx}), \sigma_{xy}=(0,0,z_{xy}), \sigma_{yy}=(0,0,z_{yy})$,
$$L=\sigma_{xx}\cdot \vec{n}=
\frac{z_{xx}}{\sqrt{1+z_x^2+z_y^2}}, M=\sigma_{xy}\cdot \vec{n}=
\frac{z_{xy}}{\sqrt{1+z_x^2+z_y^2}}, N=\sigma_{yy}\cdot \vec{n}=
\frac{z_{yy}}{\sqrt{1+z_x^2+z_y^2}}.\tag{2}$$
Combining $(1)$ and $(2)$, we get
$$K=\frac{LN-M^2}{EG-F^2}=\frac{z_{xx}z_{yy}-z_{xy}^2}{(1+z_x^2+z_y^2)^2}$$
as required. 
