How can I solve this integral with using Residues $$\int_{|z| = 1}\frac{1}{(1-3z)(1-2z)^2}dz$$
z here is a complex number calculus 

Set $\mathcal{D}=\{z\in\mathbb{C}\,:\, |z|\le1\}$. It is clear $z=\frac{1}{2}\in \mathcal{D}$ and $z=\frac{1}{3}\in \mathcal{D}$. We have
$$\text{Res}_{z=\frac 12}f(z)=\lim_{z\to\frac12}\frac{d}{dz}\left(\left(z-\frac 12\right)^2\frac{1}{(1-3z)(1-2z)^2}\right)=\frac{1}{4}\lim_{z\to\frac12}\frac{3}{(1-3z)^2}=3$$
and
$$\text{Res}_{z=\frac 13}f(z)=\lim_{z\to\frac13}\left(\left(z-\frac 13\right)\frac{1}{(1-3z)(1-2z)^2}\right)=-3$$
 By application of Reside theorem, we have
$$\oint_{|z|=1}\frac{1}{(1-3z)(1-2z)^2}dz=2\pi\text{i}(-3+3)=0$$
