 Solution

The denominator vanishes at $z_k=1/(2\pi ik)$ for every nonzero integer $k$. These points approach zero, the denominator has a simple zero at each, and $\sin z_k\ne 0$ because a nonzero purely imaginary number is not a zero of sine. Hence there are poles arbitrarily close to zero. The origin is a nonisolated singularity, so it has no residue defined by a Laurent series on a punctured disc. In particular it must not be classified as an isolated essential singularity merely because $e^{1/z}$ has one.