Let $M\subseteq \mathbb{C}$ finite set and $f:\: \mathbb{C} \setminus M \longrightarrow \mathbb{C}$ be a holomorphic function. Consider $g(z)=\frac{1}{z^{2}}f\left(\frac{1}{z}\right)$. The quiestion is: What is the $\mbox{Res}(g, 0) $? Write $\mbox{Res}(g, 0) $ in terms of the $\mbox{Res}(f, z) $.
Remark: I know that $$\mbox{Res}(g, 0)= \frac{1}{2\pi i}\int _{\left|z\right|=\varepsilon} \frac{f\left(\frac{1}{z}\right)}{z^{2}}dz.$$ Also I know that I can choose $\varepsilon>0$ small enough such that $g$ is holomorphic in $B_{\varepsilon}(0)\setminus \left\{0\right\}$. But from this last I don't know write $\mbox{Res}(g, 0) $ in terms of the $\mbox{Res}(f, z) $.
Write the Laurent series $$f(z) = \sum_{n=-\infty}^{+\infty} a_n z^n.$$ Hence $$g(z) = \sum_{n=-\infty}^{+\infty} a_n z^{-2-n}.$$
The residue is the coefficient at the rank $\frac{1}{z}$.
Thus $\mbox{Res}(f, 0) = a_{-1} = \mbox{Res}(g,0)$.
If you want to do that with integrals, a change of variable $w = \frac{1}{z}$ will give you the same answer.