I understand that the point \begin{equation} z=0 \end{equation}
is an obvious singularity, but since it's also a root of multiplicity 2 of the nominator the residue of the given function at 0 will be 0.
As for the other singularities we have
\begin{equation} \cosh z=1 \end{equation} or
\begin{equation} \cos(iz)=1 \end{equation}
and finally we find the singularities
\begin{equation} iz=2n\pi \end{equation} or \begin{equation} z_n=-i2n\pi \end{equation} for any integer.
As I have said, the residue at 0 is 0, but the other singularites are all simple poles, and therefore the residue at those points will be \begin{equation} \text{Res} f(z_n)=\lim_{z\to z_n} \frac{(z-z_n)z^2}{\cosh z-1} \end{equation} This is a 0/0 case and we can use L'Hospital's rule, and we get \begin{equation} \lim_{z\to z_n} \frac{3z^2-2z_nz}{\sinh z} \end{equation} Now this is no longer in indeterminate form, but we have \begin{equation} a/0 \end{equation}
Clearly I must have done something wrong, but I cannot seem to grasp what that something is. Any help will be greatly appreciated.
$z_n$ is a pole of order $2$ since $e^{z}+e^{-z}-2=(e^{z/2}-e^{-z/2})^{2}$. So you have to multiply by $(z-z_n)^{2}$ and compute the derivative at $z=z_n$.