How to evaluate residue of $\cot z/z^4$ at $z=0$?

The function has a pole of order $5$ at zero, so it isn't defined there.

Following @PedroTamaroff's hint and using that $\lim_{z\to 0}\frac{\sin z}{z} = 1$ : $$ \frac{\cot z}{z^4} = \frac{\cos z}{z^4\sin z}= \frac{z\cos z}{\sin z}\frac{1}{z^5} $$ Observe that the first fraction is holomorphic in a neighbourhood of $0$. Use Cauchy's theorem to finish.

Well,

\begin{align}\cot z = \frac{\cos z}{\sin z} &= \frac{1 - \frac{z^2}{2!} + \frac{z^4}{4!} + O(z^6)}{z - \frac{z^3}{3!} + \frac{z^5}{5!} + O(z^7)}\\ & = \frac{1}{z}\cdot \left(1 - \frac{z^2}{2!} + \frac{z^4}{4!} + O(z^6)\right)\cdot \left(1 + \frac{z^2}{3!} - \frac{z^4}{5!} +\frac{z^4}{3!3!}+ O(z^6)\right)\\ &= \frac{1}{z}\cdot \left(1 + \left(-\frac{1}{2!} + \frac{1}{3!}\right)z^2 + \left(-\frac{1}{2!3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{3!3!}\right)z^4 + O(z^6)\right).\end{align}

Therefore, the coefficient of $\frac{1}{z}$ in the Laurent expansion of $\frac{\cot(z)}{z^4}$ is

$$-\frac{1}{2!3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{3!3!} = -\frac{1}{45}.$$

Hence, $$\text{Res}_{z = 0}\frac{\cot(z)}{z^4} = -\frac{1}{45}.$$