$f$ has an essential singularity in $z_0$. What about $1/f$?

$f$ has an essential singularity at $z_0$ if and only if neither of the following hold:

(1) $\displaystyle \lim_{z \rightarrow z_o} f(z) \in \mathbb{C}$ (removable singularity)

(2) $\displaystyle \lim_{z \rightarrow z_o} |f(z)| = +\infty$ (pole)

Note that if $1/f$ satisfies (1) then $f$ satisfies (2) or (1) depending on whether the limit is 0 or not and if $1/f$ satisfies (2) then $f$ satisfies (1). So if $f$ has an essential singularity then so must $1/f$.

Your two examples essentially span all the possibilities. By the Big Picard Theorem, we know that $f$ assumes all but one value in $\Bbb{C}$ infinitely often, in any neighborhood of $z_0$.

If the missed value is $0$, then $\frac{1}{f}$ has an isolated singularity at $z_0$, and it must clearly be essential (since otherwise $f$ itself would have a pole or removable singularity).

If the missed value is not $0$, or if there is no missed value, then $\frac{1}{f}$ will have a sequence of poles that converges to $z_0$, and hence the singularity at $z_0$ will not be isolated. So, thinking of $\frac{1}{f}$ as a holomorphic function on its largest possible domain (which will omit some sequence of points converging to $z_0$), the singularity at $z_0$ is technically unclassifiable.

On the other hand, even in this case you can regard $\frac{1}{f}$ as a meromorphic function from $\Omega \backslash \{z_0\}$ to the extended complex plane $\hat{\Bbb{C}}$, and when considered as such it will have an essential singularity at $z_0$. This is a natural enough thing to do that I suspect most people would be happy just saying "$\frac{1}{f}$ has an essential singularity at $z_0$."

So we are given essential singularity of $f$ in $z_0$.

If the function $g=1/f$ have a pole in $z_0$ of order $m$, then $f=1/g$ will have a removable singularity. In particular this will be zero of order $m$. Contradiction, hence $z_0$ is an essential singularity of $1/f$.