Prove that the Gaussian curvature of such a surface is constant. 
The first fundamental (FF) form of a surface is given by
$$ds^2 = \frac{du^2}{(u^2+v^2+c^2)^2} + \frac{dv^2}{(u^2+v^2+c^2)^2}$$
Prove that the Gaussian curvature of such a surface is constant

I was trying to find the coefficient of the second fundamental (SF) form first. From the condition we have
$$s_u\cdot s_u = s_v\cdot s_v = \frac{1}{(u^2+v^2+c^2)^2}$$
For the (SF), I have to find $s_{uu}, s_{uv}$ and $s_{vv}$ but with the dot products given I'm not sure how to do it. I tried to differentiate w.r.t $u$ the dot product $s_u^2$ which gave
$$2\cdot s_u \cdot s_{uu} = -\frac{8(u^2+v^2+c^2)u}{(u^2+v^2+c^2)^3}$$
However, how can I extract $s_{uu}$ out of here? Or is there any other approach?
 A: There is no such thing as second fundamental form in this situation! To define second fundamental form, you require that $S$ is a regular surface in $\mathbb R^3$. But in your description, you are given only the first fundamental form and not a regular parametrization (using your notation, $s$).
So you will have to calculate the Gaussian curvature using only the first fundamental form, which can be done using (e.g.)
$$\tag{1}K = -\frac{1}{2\sqrt{EG}}\left( \frac{\partial}{\partial u} \frac{E_u}{\sqrt{EG}} +  \frac{\partial}{\partial v} \frac{G_v}{\sqrt{EG}} \right),  $$
since in your case $F = 0$ (that is, no $dudv$ terms in the first fundamental form)
Remark 1: As suggested in the comment, the given first fundamental form can be realized as the one formed by stereographic projection from the sphere of radius $1/2c$: let $s : \mathbb R^2\to \mathbb R^3$ be given by
$$ s(u, v) = \left( \frac{u}{u^2+v^2+c^2}, \frac{v}{u^2+v^2+c^2}, \frac{u^2+v^2-c^2}{2c(u^2+v^2+c^2)}\right).$$
Then
$$ s_u = \left(-\frac{2u^2}{(u^2+v^2+c^2)^2}+\frac{1}{u^2+v^2+c^2}, -\frac{2uv}{(u^2+v^2+c^2)^2}, \frac{2c u}{(u^2+v^2+c^2)^2} \right)$$
which gives
$$s_u\cdot s_u = \frac{1}{(u^2+v^2+c^2)^2}.$$
Similarly one checks $s_u \cdot s_v = 0$ and $s_v\cdot s_v = s_u\cdot s_u$. Thus $s$ gives the first fundamental form. Now you can use this explicit $s$ to calculate the second fundamental form, and from there, the Gaussian curvature.
Remark 2: Strictly speaking, there is a gap in Remark 1: there might be another $s_2$ which induces the same first fundamental form, but different second fundamental form. Indeed, the regular parametrizations
$$ s_1 (u, v) = (u, v, 0), \ \ \ s_2 = (u, \cos v, \sin v)$$
induce the same first fundamental form: $du^2 + dv^2$, but $s_1$ has identically zero second fundamental form while $s_2$ isn't.
So one might wonder if that would gives us different Gaussian curvature. It is the Gauss's Theorema Egregium that tells you that this is not the case: the Gaussian curvature does not depend on the second fundamental form. So with Gauss's Theorema Egregium, one can just sticks with the parametrization given in Remark 1 and starts the calculation (well, but then it would be easier to just use (1), which is also a consequence of the same theorem).
