I passed through this answer here and I don't understand the following. We are at the moment:
If $u^2+v^2 \neq 0$, then $\dfrac{\partial v}{\partial y}=0$ which also gives us $\dfrac{\partial v}{\partial x}=0$. Now by Cauchy-Riemann, we also get that $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial y}=0$. Hence, $f'(z) = 0$ which gives us that $f$ is constant.
Why we get "$\dfrac{\partial v}{\partial y}=0$ which also gives us $\dfrac{\partial v}{\partial x}=0$? I don't see this...
There is a comment in the answer to the question you linked that answers your question. The original question was about proving that for a holomorphic function $f$, $|f|$ constant implies $f$ constant. Now if $|f(z)|$ is constant, so is $|f(iz)|$, and running the same argument that showed $\frac{\partial v}{\partial y}=0$ on this rotated function gives $\frac{\partial v}{\partial x}=0$.