We consider $$\sigma (u,v)=(f(u)\cos v, f(u)\sin v, g(u))$$ Picking $u=\theta , v=\phi , f(\theta )=\cos \theta , g(\theta )=\sin \theta$ we get that the first fundamental form is $$d\theta^2+\cos^2 \theta d\phi^2$$
We consider the reparametrization $\tilde{\sigma}(u, v) = \sigma (\psi(u), v)$.
I want to show that the reparametrization is conformal.
I have done the following:
The first fundamental form of $\tilde{\sigma}$ is $$(\psi '(u))^2du^2+\cos^2(\psi(u))dv^2$$
So that the reparametrization is conformal it must stand $$(\psi'(u))^2=\cos^2 (\psi (u)) \Rightarrow \psi '(u)=\pm \cos (\psi (u)) \Rightarrow \frac{d\psi}{du}=\pm \cos \psi \Rightarrow \frac{1}{\cos\psi}d\psi=\pm du$$
We have that
$$\int \frac{1}{\cos \psi}d\psi=\ln \sqrt{\frac{1+\sin \psi}{1-\sin \psi}}+C=\ln \sqrt{\frac{(1+\sin \psi)(1-\sin \psi)}{(1-\sin \psi)^2}}+C \\ =\ln \sqrt{\frac{1-\sin^2\psi}{(1-\sin \psi)^2}}+C=\ln \sqrt{\frac{\cos^2\psi}{(1-\sin \psi)^2}}+C=\ln \frac{\cos\psi}{1-\sin \psi}+C$$
Therefore, we have the following: $$\int \frac{1}{\cos \psi}d\psi=\pm \int du \Rightarrow \ln \frac{ |\cos\psi |}{1-\sin \psi}+C=\pm u \Rightarrow \ln \frac{ |\cos\psi |}{1-\sin \psi}=\pm u-C\\ \Rightarrow \frac{ |\cos\psi |}{1-\sin \psi}=\tilde{C}e^{\pm u}$$
Therefore, the smooth function $\psi$ that we are looking for is given by the relation $\frac{ |\cos\psi |}{1-\sin \psi}=\tilde{C}e^{\pm u}$, for $\cos\psi\neq 0$.
Is this correct? Could we get a specific formula for $\psi$ ?
How can we show that $\tilde{\sigma}$ is the Mercator parametrization $$\sigma (u,v)=(\text{sech } u \cos v, \text{sech } u \sin v, \tanh u)$$ ?
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EDIT:
The first fundamental form of $\tilde{\sigma}$ is $$(\psi '(u))^2du^2+\cos^2(\psi(u))dv^2$$
So that the reparametrization is conformal it must stand $$(\psi'(u))^2=\cos^2 (\psi (u)) \Rightarrow \psi '(u)=\pm \cos (\psi (u)) \Rightarrow \frac{d\psi}{du}=\pm \cos \psi \Rightarrow \frac{1}{\cos\psi}d\psi=\pm du$$
We have that
$$\int \frac{1}{\cos \psi}d\psi=\frac{1}{2}\ln \frac{1+\sin \psi}{1-\sin \psi}+C=\text{arctanh } (\sin\psi)+C$$
Therefore, we have the following: $$\int \frac{1}{\cos \psi}d\psi=\pm \int du \Rightarrow \text{arctanh } (\sin\psi)+C=\pm u \Rightarrow \text{arctanh } (\sin\psi)=\pm u-C\\ \Rightarrow \sin\psi=\tanh (\pm u-C) \Rightarrow \psi=\text{arcsin } (\tanh (\pm u-C))$$
Therefore, the reparametrization is $$\tilde{\sigma}(u, v) = \sigma (\psi(u), v) \\ =(\cos (\text{arcsin } (\tanh (\pm u-C)) )\cos v, \cos (\text{arcsin } (\tanh (\pm u-C)) )\sin v, \sin (\text{arcsin } (\tanh (\pm u-C)) )) \\ =(\cos (\text{arcsin } (\tanh (\pm u-C)) )\cos v, \cos (\text{arcsin } (\tanh (\pm u-C)) )\sin v, \tanh (\pm u-C)) )\\ =(\sqrt{1-\tanh^2 (\pm u-C)}\cos v, \sqrt{1-\tanh^2 (\pm u-C)}\sin v, \tanh (\pm u-C)) )=(\text{sech } (\pm u-C)\cos v, \text{sech } (\pm u-C)\sin v, \tanh (\pm u-C)) )$$
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How do we get rid of $C$ and $\pm$ ?
No, it's not correct. Remember that $\theta = \psi(u)$, so you need $\psi'(u) = \pm \cos(\psi(u))$.
EDIT: Now solve this separable differential equation.