Problem
Suppose $z,f$ are two meromorphic functions on a compact Riemann surface $M$, whose meromorphic function field is $\mathbb C(M)=\mathbb C(z,f)$, where $\mathbb C(M)$ is a finite extension of $\mathbb C(z)$. Let $A$ be the set of poles of $z,f$ along every $p\in M$ such that $dz(p)=0$ or $df(p)=0$. Then for each $p,q\in M\setminus A$, $p\neq q$, we have $(z(p),f(p))\neq(z(q),f(q))$.
It's lemma 21.2 on Wu Hongxi's book on compact Riemann surfaces. Suppose that $z(p)=z(q)$ and $f(p)=f(q)$. It's a corollary of Riemann-Roch formula that if $p\neq q$, we can choose $h\in\mathbb C(M)$ such that $h(p)\neq h(q)$. However, if $h=\alpha(z,f)/\beta(z)$ where $\alpha\in\mathbb C[X,Y]$ and $\beta\in\mathbb C[X]$ are two polynomials, we have $h(p)=\alpha(z(p),f(p))/\beta(z(p))=\alpha(z(q),f(q))/\beta(z(q))=h(q)$. Contradiction! Everything is okay if $\beta(z(p))=\beta(z(q))\neq0$. When $\beta(z(p))=\beta(z(q))=0$, I cannot see any evidence for the preceding equality. I need some hint on this step.
Background
Such a lemma is used to show that a compact Riemann surface $M$ could be holomorphically mapped onto a plane algebraic curve, and by removing finitely many points on $M$, the map is injective. In fact, the totality of these finitely many points could be chosen as $A$ above.
Then we can show that every compact Riemann surface could be holomorphically embedded in $P^2(\mathbb C)\times P^1(\mathbb C)\subseteq P^5(\mathbb C)$ (the Segre embedding). It's not the canonical way to show that any compact Riemann surface could be embedded in $P^3(\mathbb C)$ though.