I was trying to find out the residue of the function: $$f(z)=\frac{\pi\cot(\pi z)}{z^2}$$
It is evident that we have $z=0$ as a pole of order $3$.
So we have: $$\operatorname*{Res}_{z = 0}f(z)=\frac{1}{2}\lim_{z \to 0}\frac{d^2}{dz^2}\left(z^3\times \frac{\pi\cot(\pi z)}{z^2}\right)$$ So we get:
$$\operatorname*{Res}_{z=0}f(z)=\frac{\pi}{2}\lim_{z \to 0}\frac{d}{dz}\left(-\pi z \csc^2 \pi z+\cot(\pi z)\right)$$
But it’s tedious to continue from here. Is there any alternate way?
The formula you’re applying is perhaps best viewed as a bookkeping device to extract the desired coefficient of $z^{-1}$ of the Laurent series of $f$ at $z=0$. In the present case, it seems easier to work directly with the series. From Laurent series for $\cot (z)$, we have
$$ \cot z=\frac1z-\frac z3+O\left(z^3\right)\;, $$
so
$$ \frac{\pi\cot(\pi z)}{z^2}=\frac1{z^3}-\frac{\pi^2}3\cdot\frac1z+O(z)\;, $$
from which you can read off that the residue at $z=0$ is $-\frac{\pi^2}3$.
$$Res(f,0)=\frac{1}{2}\lim_{z \to 0}\frac{d^2}{dz^2}\left(z^3\times \frac{\pi\:\cot(\pi z)}{z^2}\right)$$
$=lim_{z\rightarrow 0}\frac{d^2}{dz^2}(πz\times \frac{\pi cos(π z)}{sin(πz)})$
$= -π^2/2$ as $\lim_{z\rightarrow 0}\frac{sin(πz)}{πz}=1$
