Since $|z| > 1$, with finding a convergent series in mind I wanted to write an equivalent closed-form expression of $f$ in terms of $\frac 1z$ because $|\frac 1z| < 1$. So I tried to factor out $\frac 1{z^2}$ and writing $f$ as follows: $$ f(z) = \frac 1{z^2}\frac 1{1 - (-\frac 1z - \frac 1{z^2})}. $$ However, $|-\frac 1z - \frac 1{z^2}| \le \frac 1{|z|}+\frac 1{|z|^2} < 1+1=2$, so I was unable to bound $|-\frac 1z - \frac 1{z^2}|$ by $1$.
What is a suitable expression of $f$ that I would need so that I can finally rewrite it in terms of a convergent series?
Should I consider splitting into two cases: $1 < |z| \le 2$ and $|z| > 2$? It turns out that if $|z| > 2$, then $|-\frac 1z - \frac 1{z^2}| \le \frac 1{|z|}+\frac 1{|z|^2} < \frac 12 + \frac 12 = 1$.
Since $|z|>1$ we can consider $|w|=\left|\dfrac{1}{z}\right|<1$, then \begin{align} \dfrac{1}{z^2+z+1} &= \dfrac{w^2}{1+w+w^2} \\ &= \dfrac{w^2(1-w)}{1-w^3} \\ &= (w^2-w^3)(1+w^3+w^6+\cdots) \\ &= \dfrac{1}{z^2}-\dfrac{1}{z^3}+\dfrac{1}{z^5}-\dfrac{1}{z^6}+\dfrac{1}{z^8}-\dfrac{1}{z^9}+\cdots \end{align}