I am having some difficulty calculating the residue of $\frac{1}{z^2 \sin z}$ at $z = 0$. From what I can tell, we are dealing with an essential singularity here and so the problem becomes that of finding the Laurent expansion and this is where I am stuck. I tried rewriting the term in the following way
$\frac{1}{z^2 \sin z} = \frac{1}{z^2}(\frac{1}{\sin z}) = \frac{1}{z}-\frac{z}{3!}+\frac{z^2}{5!}-\ldots=\sum_{n=1}^\infty (-1)^{n-1}\frac{z^{2n}}{(2n-1)!}$
and I am stuck here on what to do next, any hints?
No, it is not an essential singularity. It is a pole of order $3$, since$$\lim_{z\to0}z^3\frac1{z^2\sin(z)}=\lim_{z\to0}\frac z{\sin(z)}=1\neq0.\tag1$$
On the other and, your function is an odd one, and therefore its Laurent series is of the type$$\frac{a_{-3}}{z^3}+\frac{a_{-1}}z+a_1z+\cdots$$Actually, it follows from $(1)$ that $a_{-3}=1$. And\begin{align}\operatorname{res}_{z=0}\left(\frac1{z^2\sin(z)}\right)&=a_{-1}\\&=\lim_{z\to0}z\times\left(\frac1{z^2\sin(z)}-\frac1{z^3}\right)\\&=\lim_{z\to0}\frac{z-\sin(z)}{z^2\sin(z)}\\&=\frac16.\end{align}