$\displaystyle\int_{|z|=3}{\frac{9z^5}{(z-2)(1+2z)^2(1-3z)^3}}dz$
I tried evaluating this directly and it has turned into a nightmare of a problem. I was thinking this is more easily solved using the residue at infinity. Are my intuitions correct? If so I'm not really sure how to go about doing so.
Yes, your intuition is correct. Note that $$\operatorname*{Res}_{z=\infty} f(z) +\frac1{2\pi i}I=0$$ where $I$ is your integral that happens to include all poles.
Recall $$\operatorname*{Res}_{z=\infty} f(z)=-\operatorname*{Res}_{z=0}\frac1{z^2}f\left(\frac{1}{z}\right)$$
simplifying gives
$$\operatorname*{Res}_{z=\infty}f(z) = -\operatorname*{Res}_{z=0}\frac1{z^2}\frac{9z}{(1-2z)(z+2)^2(z-3)^3}=-\frac9{1\cdot4\cdot (-27)}$$
Thus $$\operatorname*{Res}_{z=\infty} f(z) =\frac1{12}$$ and $$I= -\frac{\pi i}{6}$$.