Let the sphere be parametrized by
$$\{\cos (u) \cos (v),\sin (u) \cos (v),\sin (v)\}$$
where $0\leq u \leq 2\pi$ and $-\pi/2 \leq v \leq \pi/2$.
I wanted a projection of the sphere on $\mathbb{R}^2$
$$\{\cos(u)r(v),\sin(u)r(v)\}$$
where $r(v)$ os an increasing function with $r(-\pi/2)=0$, and such that, given $v$, the circumference over $u$ in the 2 parametrizations have a ratio equal to the ratio of arclength derivatives wrt. $v$. These ratios are:
$$\frac{2 \pi r(v)}{2 \pi \cos (v)}=\frac{r'(v)}{1} $$
which implies $r(v)=ke^{2 \tanh ^{-1}(\tan(v/2))}$.
What is the thisprojection of the sphere onto $\mathbb{R}^2$ called?
Note that your $r(v)$ can be simplified to $$r(v)={1+\tan(v/2)\over 1-\tan(v/2)}={\cos v\over 1-\sin v}={\sqrt{x^2+y^2}\over 1-z}$$ (I've put $k=1$). It turns out that the map you constructed is the good old stereographic projection. This does not come as a surprise: The map is set up in such a way that at each point two specific orthogonal tangent directions are mapped to orthogonal tangent directions, and it is taken care that the linear stretching is the same in both these directions. It follows that the map is conformal; furthermore it is rotationally symmetric with respect to the $z$-axis.