Question: I want to express $$\sum\_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$$ in closed form.

I know that $$\pi z\cot(\pi z)=1+2z^2\sum\_{n=1}^\infty\dfrac{1}{z^2-n^2}$$ which looks close, but I don’t know how to use it.

Answer: 
First rewrite it as:

$$\frac{1}{z^3} + \sum\_{n=1}^\infty \left(\frac{1}{z^3-n^3} + \frac{1}{z^3+n^3}\right) = \frac{1}{z^3} +\sum\_1^\infty \frac{2z^3}{z^6-n^6}$$

so we only need to find:

$$\sum\_{n=1}^\infty \frac{1}{z^6-n^6}$$

Now, if $\omega$ is a primitive cube root of unity, there are functions $a(u),b(u),c(u)$ such that for real $u\neq v$:

$$\frac{1}{u^3-v^3} = \frac{a(u)}{u-v} +\frac{b(u)}{\omega u -v} + \frac{c(u)}{\omega^2u-v}$$

Setting $u=z^2$ and $v=n^2$, and finding a solution $\zeta$ so that $\zeta^2=\omega$, then we have:

$$\frac{1}{z^6-n^6} = \frac{a(z^2)}{z^2 - n^2}+\frac{b\left((\zeta z)^2\right)}{(\zeta z)^2-n^2} + \frac{c\left((\zeta^2 z)^2\right)}{(\zeta^2 z)^2 -n^2}$$

So we can use your first result to find:

$$\sum\_{n=1}^\infty \frac{1}{z^6-n^6} = a(z^2)f(z) + b(\zeta^2z^2)f(\zeta z) + c(\zeta^4 z^2)f(\zeta^2 z)$$
where $$f(z)=\sum\_{n=1}^\infty \frac{1}{z^2-n^2} = \frac{\pi z\cot(\pi z)-1}{2z^2}$$
That seems a bit messy, and requires jumping into complex numbers...
Assuming you are referring to the book by Ahlfors, here is a method relying on it:

Starting with a partial fraction decomposition w.r.t. $n$ we find

$$\frac{1}{z^3-n^3}=\frac{ \rho\_3^0}{3z^2} \frac{1}{ \rho\_3^0 z-n}+\frac{\rho\_3^1}{3z^2} \frac{1}{ \rho\_3 z-n}+\frac{\rho\_3^2}{3z^2} \frac{1}{ \rho\_3^2 z-n} $$ where $\rho\_3=\exp(2 \pi i/3)$. Using the formula $$ \pi \cot( \pi z)= \lim\_{m \to \infty} \sum\_{n=-m}^m \frac{1}{z-n} $$
from page 189 three times should give you the desired result. I will add that since the series converges in the usual sense, it also converges w.r.t. this "principal value summation".