The task says:
Show that if some function $\;f(z)={1\over g(z)}$, where $g\not\equiv0$ is an entirely analytic function, then the isolated singularities of $\;f$ are exactly zeros of $g$ and namely are the poles, that have the same orders as the orders of the corresponding zeros of $g$.
As I understand it correctly (relying on what Wikipedia says) the notation $g\not\equiv0$ means that $g$ is never equal to $0$, that means for any $z$, which would be the same if we said that $g$ doesn't have zeros. But the taks asks us to prove that these not existing zeros are the singularities of $\;f(z)$.
Is there something wrong with my knowledge of mathematical notation or with the taks itself?
The notation means that the function $g$ is not identically zero. This means that $g(z) = 0$ for all $z$ does NOT hold.