Evaluating $I=oint frac{cos(z)}{z(e^{z}-1)}dz$ along the unit circle

This integral can be expressed with help of the residue theorem: $$I=oint frac{cos(z)}{z(e^{z}-1)}dz=2pi i sum_{k}; mathrm{Res}(f,a_k)$$
There is only one removable singularity at $z=0$ and you should be able to find the redsidue of $f$ at $z=0$, just find the coeficient at $z^1$ of the following series expansion of $z^2 f(z)$: $$ z^2 frac{cos(z)}{zleft(e^{z}-1right)}=frac{z cos(z)}{e^{z}-1}=1-frac{z}{2}+O(z^2)$$ i.e. $$mathrm{Res}left(frac{cos(z)}{zleft(e^{z}-1right)},0right)=-frac{1}{2}$$ and so $$I=oint frac{cos(z)}{z(e^{z}-1)}dz=- i pi$$

For a bit more involved calculation based on the Cauchy integral theorem see e.g. this answer or using simply the residue theorem see this one.