I am looking at the following exercise:
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Could you give me some hints how we could show that?
Do we use the matrix of the Weingarten map with respect to the basis $\{\sigma_u,\sigma_v\}$ of the tangent plane? But how exactly?
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EDIT1:
The matrix of the Weingarten map with respect to the basis $\{\sigma_u,\sigma_v\}$ of the tangent plane is $$\mathcal{F}_I^{-1}\mathcal{F}_{II}= \begin{pmatrix} \cos^2 v & 0 \\ 0 & 1 \end{pmatrix}^{-1}\begin{pmatrix} -\cos^2 v & 0 \\ 0 & -1 \end{pmatrix}=\frac{1}{\cos^2 v}\begin{pmatrix} -\cos^2 v & 0 \\ 0 & -\cos^2 v \end{pmatrix}=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$$
We have that $$\mathcal{F}_I^{-1}\mathcal{F}_{II}= \begin{pmatrix} a & c \\ b & d \end{pmatrix}$$ where $-\textbf{N}_u=a\sigma_u+b\sigma_v$ and $-\textbf{N}_v=c\sigma_u+d\sigma_v$.
Therefore, we get $-\textbf{N}_u=-\sigma_u\Rightarrow \textbf{N}_u=\sigma_u \Rightarrow \textbf{N}=\sigma+A(v)$ and $-\textbf{N}_v=-\sigma_v \Rightarrow \textbf{N}_v=\sigma_v \Rightarrow \textbf{N}=\sigma+B(u)$.
From the relations $\textbf{N}=\sigma+A(v)$ and $\textbf{N}=\sigma+B(u)$, we see that $A(v)=B(u)$ must be a constant. So $\textbf{N}=\sigma+\tilde{C}$.
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EDIT2:
Could you give me some hints how we could find a parametrization of $S^2$ with the above first and second fundamental forms?
Let me give an explicit answer: Write $M$ for your surface.
First, the $2-$nd fundamental form is typically defined as
$II(u,v) := - I(d \nu (u) , v)$
where $\nu : M \rightarrow S^2$ is the Gauss map, i.e. the map that assigns to each point $p$ the unit normal vector to the tangent plane of $p$. (Note that I had a different sign in my comment. I was used to a different convention, but let us stick with how it is on wikipedia)
The first and second fundamental forms in your example fulfill the relationship $II = - cos(v)du^2 - dv^2 = - I$, hence $d \nu$ must be the identity map. This means in particular, that $\nu$ is a local isometry, i.e. around each point $p$ there exists an open neighbourhood $U$ of $p$, such that $\nu |_U$ is an isometry.
PS: A small remark is that the proof of Mindings theorem for surfaces of positive curvature goes in much the same way.
PPS: What I call $\nu$, you denoted by $N$ in your edit. It's the same function.