Let $f(z) = z + a_{2}z^{2} + a_{3}z^{3} + \ldots $ be the Taylor series of $f$ which is valid for $ \vert z \vert < 1$.
I want to prove that the Laurent series of $g(z) = 1/f(1/z)$ is of the form $g(z) = z + b_{0} + b_{1}/z + b_{2}/z^{2} + \ldots $ which is valid for $ \vert z \vert > 1$.
I found that $z^{2}f(1/z) = z + a_{2} +a_{3}/z + \ldots$, but how can I associate it with $1/f(1/z)$?
One way to see it is as follows. Let's say $g(z)=\sum\limits_{n=-\infty}^\infty b_n z^{-n}$ (you already know that it is analytic). Then, $$\lim_{z\to\infty}\frac{g(z)}{z}=\lim_{z\to\infty}\frac{1}{zf(1/z)}=\lim_{z\to\infty}\frac{1}{1+a_2/z+a_3/z^2+\dots}=1$$ But this can only be the case for $b_{-1}=1$ and $b_{-n}=0$ for $n\geq2$ since $$\frac{g(z)}{z}=\underbrace{\left(\sum\limits_{n=0}^\infty b_n z^{-n-1}\right)}_{\to 0 \text{ when } z\to\infty}+b_0+b_{-1}z+b_{-2}z^2+\dots$$
Another equivalent way to see it is the following one. Notice that $g(1/z)=\sum\limits_{n=-\infty}^\infty b_n z^{n}$. Then, it can be analogously checked that $\lim_{z\to0}zg(1/z)=\lim_{z\to0}z/f(z)=1$ which necessarily implies that $b_{-1}=1$ and that $b_{-2}=b_{-3}=\dots=0$.