Short proof that a surface has constant Gaussian curvature Let $M \subset \Bbb R^3$ be a regular surface with a parametrization where the induced metric is $$ds^2=\frac{du^2+dv^2}{(u^2+v^2+c)^2}$$
Where $c>0$ is constant. Prove that the Gaussian curvature of $M$ is constant.

I succeeded to solve it by computing the Riemannian curvature by the definition, but it was very long and not elegant. Is there a simpler way to calculate it, by using the direct definition of Gaussian curvature (i.e using a parametrization)?
 A: The curvature of $$\frac{du^2+dv^2}{h(u,v)^2}$$ is given by $$K = h\,\triangle h - \|\nabla h\|^2.$$Here $h(u,v)=u^2+v^2+c$ gives that $$K = 4(u^2+v^2+c) -(4u^2+4v^2) = 4c.$$The proof of this formula is a direct application of Gauss' formula $$K = \frac{-1}{\sqrt{EG}}\left(\left(\frac{(\sqrt{G})_u}{\sqrt{E}}\right)_u+\left(\frac{(\sqrt{E})_v}{\sqrt{G}}\right)_v\right)$$for orthogonal parametrizations.
A: I will give an answer using solely differential forms which I believe gives a more elegant solution as it rewrites the above formulae using less variables; hopefully you are familiar with them. In my opinion (not necessarily right or more instructive), it is easier to remember this flow of ideas than the formula-based approach. 
The induced metric (first fundamental form) can be written as
$$I=\theta_1^2+\theta_2^2,$$
where in your case you have $\theta_1=\dfrac{du}{u^2+v^2+c}$ and $\theta_2=\dfrac{dv}{u^2+v^2+c}$.
Also, we know that the Gaussian curvature appears in the Gauss equation
$$d\omega_{12} = K\theta_1\wedge\theta_2,\tag{1}\label{Gauss}$$
where $\omega_{12}$ is the $\textit{connection form}$ satisfying the structure equations
$$d\theta_1 + \omega_{12}\wedge\theta_2=0$$
$$d\theta_2 - \omega_{12}\wedge\theta_1=0$$
Computing the exterior derivatives you get $d\theta_1=\dfrac{2v\;du\wedge dv}{(u^2+v^2+c)^2}$ and $d\theta_2=-\dfrac{2u\;du\wedge dv}{(u^2+v^2+c)^2}$. Using these along with the two equations above you get 
$$\omega_{12}=\dfrac{-2v\;du+2u\;dv}{u^2+v^2+c},$$
which replaced in (1) gives you $\dfrac{4c\;du\wedge dv}{(u^2+v^2+c)^2}=K\theta_1\wedge\theta_2$ from which you can conclude $K=4c$.
Hope it helps and shows a less computational way of finding out the Gauss curvature.
