McKean & Moll offer the following sketch of a proof.
Let $P(x)=a_0(x)-f_0b_0(x)$ in $\mathbb C(f_0)[x]$ where $f_0(z)=\dfrac{a_0(z)}{b_0(z)}$ is in $\mathbb C(z)$. Then evidently, $\deg P=\#$ poles of $f_0=$ topological degree of $f_0$ as it extends to a map from $\mathbb P^1\to\mathbb P^1$. Hence, $d=[\mathbb C(z):\mathbb C(f_0)]\leq\deg P=$topological degree of $f_0$. On the other hand, let $P_0=x^d+s_1x^{d-1}+\dots s_d$ be the minimal polynomial of $z$ over $\mathbb C(f_0)$. Clearly (or it is not hard to check that) only finitely many $c\in\mathbb C$ are poles of some $s_i$ or such that $f(z)=c$ has repeated roots.
So far so good, but then comes the claim I can't prove.
Claim. If $f(z)=c$ has only simple roots (and so $d_0=$topological degree of $f_0$ of them), then $x^d+s_1(c)x^{d-1}+\dots s_d(c)$ vanishes on each of those simple roots (where $s_i(c)$ is the image of $s_i$ under the map that takes $f_0\to c$ in $\mathbb C(f_0)$; the $s_i$ being rational functions of $f_0$).
I see why wherever $x^d+s_1(c)x^{d-1}+\dots s_d(c)$, we have a root of $f(z)=c$, but I don't see what the converse would follow from at all. Any hints would be greatly appreciated.