Note it's symmetric in $z$ and ${1 \over z}$, so if you know the sum for $|z| > 1$ then you have the sum for $|z| < 1$ via $f(z) = f(1/z)$.
Note that your series is ${\displaystyle \sum_{n = 0}^{\infty} {z^{-n} \over 1 + z^{-2n}}}$. Given any $z$ with $|z| > 1$, if $n$ is large enough, then $|z^{-2n}| = |z|^{-2n} < {1 \over 2}$, so that $|1 + z^{-2n}| \geq 1 - |z|^{-2n} > { 1 \over 2}$. So for such $n$ you have $\bigg|{z^{-n} \over 1 + z^{-2n}}\bigg| < 2|z|^{-n}$ So since the sum of $|z|^{-n}$ is convergent when $|z| > 1$, the original series is too.