I tried
$$\frac{1}{z^2(z-3)^2} = \frac{1}{((z-3)^2 - 6(z-3) + 9)(z-3)^2 }$$
but this doesn't lead somewhere.
Am I supposed to calculate the coefficients using
$$a_n = \frac{1}{2\pi i }\oint_\gamma \frac{f(z)dz}{(z-a)^{k+1}}$$?
This seems like too much work.
Any hints?
First of all, note that\begin{align}\frac1z&=\frac1{3+(z-3)}\\&=\frac13\times\frac1{1+\frac{z-3}3}\\&=\frac13\left(1-\frac{z-3}3+\frac{(z-4)^2}{3^2}-\frac{(z-3)^3}{3^3}+\cdots\right)\\&=\frac13-\frac{z-3}{3^2}+\frac{(z-4)^2}{3^3}-\frac{(z-3)^3}{3^4}+\cdots\end{align}Differiating both sides and multiplying by $-1$, you get$$\frac1{z^2}=\frac1{3^2}-2\frac{z-4}{3^3}+3\frac{(z-4)^2}{3^4}-\cdots$$and therefore$$\frac1{z^2(z-3)^2}=\frac1{3^2(z-3)^2}-2\frac1{3^3(z-3)}+3\frac1{3^4}-\cdots$$