Does this qualify as a Laurent series? Evaluate the Laurent series around the singularity at $ z_0 = 3$.
$$ \frac{1}{z^2(z-3)} $$

I can apply the geometric series as follows:
$$\frac{1}{z^2}\cdot\frac{1}{z-3}=-\frac{1}{3z^2}\cdot\frac{1}{1+\frac{z}{3}}=\frac{1}{3z^2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{3}\right)^n$$
though I'm not sure that if the final answer, since the Laurent series involves negative powers of $ n $.
What do you think?

For a Laurent series about the point $z_0=3$, you want the series to be in powers of $z-3$.
$$
\begin{align}
\frac1{z^2(z-3)}
&=\frac1{((z-3)+3)^2(z-3)}\\
&=\frac1{9\left(1+\frac{z-3}3\right)^2}\cdot\frac1{z-3}\\
&=\frac19\frac1{z-3}\left(1-2\frac{z-3}3+3\frac{(z-3)^2}9-4\frac{(z-3)^3}{27}+\dots\right)\\
&=\frac1{9(z-3)}-\frac2{27}+\frac{3(z-3)}{81}-\frac{4(z-3)^2}{243}+\dots\\
&=\sum_{k=0}^\infty\frac{k+1}{(-3)^{k+2}}(z-3)^{k-1}
\end{align}
$$
