$f(z) = \frac {e^{z^{2}}}{z^3}$, valid for $|z| >0$
The way I am thinking about this problem is to factor out the $e^{z^{2}}$ term as I am use to Laurent series of the form $\frac{1}{1-w}$. I am a little lost over whether this is the correct approach and if so, what is the next step, or if I'm approaching this question in the wrong way.
$$\frac{e^{z^2}}{z^3}=\frac{1}{z^3}\bigg{(}1+z^2+\frac{z^4}{2!}+\frac{z^6}{3!}+\dots\bigg{)}=\frac{1}{z^3}+\frac{1}{z}+\frac{z}{2!}+\frac{z^3}{3!}+\frac{z^5}{4!}+\dots$$
If you're wondering why the multiplication is "allowed", remember that the series converges uniformly over compact subsets of $\mathbb{C}\setminus\{0\}$.