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For $z\ne 0$, the denominator zeroes are $z_n=1/[\pi (n+\frac{1}{2})]$. Its derivative is $\sin (1/z)/z^2$, nonzero at each such point. ThusThese poles accumulate at zero. Consequently zero is a non-isolated singularity, as described by reciprocal cosine with accumulating poles. It is not an isolated essential singularity: no punctured disc about zero is a domain of holomorphy for this reciprocal function. Although $\cos (1/z)$ itself has an essential singularity at zero, taking its reciprocal introduces this accumulation of poles.

At every $z_k$, the function $\tan (1/z)$ has a simple pole. The exponential of a pole is an essential singularity, soare essential singularities of $\exp (\tan (1/z))$.

These points accumulate at $0$. Therefore no punctured neighbourhood of zero is a domain of holomorphy for the function: zero is a non-isolated singularity, specifically an accumulation point of essential singularities, rather than an isolated essential singularity. There are no other singularities in $\mathbb{C}$.

The function $1/\cos (1/z)$ has simple poles at $z=1/[\pi (n+1/2)]$, accumulating at zero. Its origin is therefore a non-isolated singularity, despite the underlying cosine of $1/z$ having an isolated essential singularity.