How to show
$f, g$ are analytic at $z_0$. Then $f/g$ is analytic at $z_0$ iff $g(z_0) ≠ 0.$
using $f, g$ are differentiable at $z_0$. Then $f/g$ is differentiable at $z_0$ iff $g(z_0) \neq 0.$
$\Rightarrow$ follows immediately. How to prove $\Leftarrow?$
I came to know "it is customary in the analytic context to replace removable singularities by their limit." But what if we exclude the custom?
$g$ is analytic at $z_0\implies g$ is continuous at $z_0\implies \lim_{z\to z_0}g(z)=g(z_0)\ne 0\implies \exists$ a neighborhood $N$ of $z_0$ such that $g(z)\ne 0~\forall~z\in N\cap$ domain of $g\implies$ $g(z)\ne 0~\forall~z\in N\cap$ domain of $g.$
$f,g$ are analytic at $z_0\implies f,g$ are differnetaible in a neighborhood $H$ of $z_0.$
So $g(z)\ne 0~\forall~z\in A=H\cap N,$ a neighborhood of $z_0.$
Consequently $f/g$ is differentiable in $A.$