There is a theorem that says rational functions in the extended complex plane are exactly the meromorphic functions.
After this, my textbook draws the corollary: "...as a consequence, a rational function is determined up to a multiplicative constant by prescribing the locations and multiplicities of its zeros and poles."
I don't see how it follows. Plus I think I have an example. Take a rational function with no zeros and two poles of multiplicity one, at 0 and 1.
At least two such functions exist:
$f(x)=\frac{1}{x} + \frac{1}{x-1}$
$g(x)=\frac{1}{x} + \frac{2}{x-1}$
The precise statement is the other way round: a function meromorphic on the whole Riemann sphere $S$ is rational.
This is completely false on $\mathbb C$: the function $e^z$ is meromorphic (even holomorphic) on $\mathbb C$, but certainly not rational on $\mathbb C$, nor anywhere else.
Now if $f,g\in Rat(S)$ are rational on the Riemann sphere $S$ and have the same zeros and poles (counted with multiplicities), then the quotient $\phi=f/g$ having neither zeros nor poles is holomorphic on the whole of $S$ and thus constant (say because $\phi$ is bounded on $S$ by compacity, hence bounded on $\mathbb C$ and Liouville's theorem applies).
"Yes,Georges, but I gave you a counterexample"
Your rational functions don't have the same zeros: $$f(\frac {1}{2})=0\neq g(\frac {1}{2})=-2 $$
A very optional remark
That meromorphic functions on $S$ are rational is the simplest example of a very profound result in algebraic geometry: Serre's GAGA principle.