= Solution
In local coordinates centred at $p$ and $f(p)$, a nonconstant rational map has the form
$$
w=z^{e_p}u(z),\qquad u(0)\ne0.
$$
The point $p$ is a <ramification point of a holomorphic map> when its <ramification index of a holomorphic map> satisfies $e_p>1$. A <branch value of a holomorphic map>, also called a branch point in the target, is a value $f(p)$ of some ramification point.
Let $B$ be the finite set of branch values and put $R=f^{-1}(B)$. For every $w\notin B$, all points of $f^{-1}(w)$ have local degree one, so the <holomorphic inverse function theorem> supplies disjoint neighbourhoods on which $f$ is biholomorphic. Compactness of the fibre lets their target neighbourhoods be intersected to one evenly covered neighbourhood of $w$. Hence
$$
\boxed{f:\mathbb C_\infty\setminus R
\longrightarrow\mathbb C_\infty\setminus B}
$$
is an unramified covering map. Here the displayed restriction requires $R=f^{-1}(B)$; if “ramification point” is reserved only for points with $e_p>1$, then the source deletion must be written $f^{-1}(B)$.
The <Monodromy theorem> says that analytic continuations along endpoint-fixed homotopic paths have the same terminal germ. Equivalently here, the <path lifting theorem> lifts a loop $\gamma$ based at $w\notin B$ from each point of $f^{-1}(w)$; taking the endpoint of each lift permutes that fibre. The permutation depends only on $[\gamma]\in\pi_1(\mathbb C_\infty\setminus B,w)$, giving the <monodromy group of a covering>.
For the stated function, make the Möbius change of coordinate
$$
t=\frac1{1-z}.
$$
Then
$$
f(z)=t^4-t^2.
$$
The critical points are $t=0,\pm1/\sqrt2,\infty$, with branch values
$$
0,\qquad-\frac14,\qquad\infty.
$$
A loop around $0$ interchanges the two roots born from $t=0$, so its branch cycle is a transposition. A loop around $-1/4$ simultaneously interchanges the two pairs that collide there, so its branch cycle is a product of two disjoint transpositions. With a suitable labelling these are
$$
(34),\qquad(13)(24).
$$
They generate a transitive group of order eight; their product is a four-cycle. Therefore the full monodromy group is
$$
\boxed{D_4\le S_4},
$$
the <dihedral group> in its action on the four vertices of a square.
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