Can there be a point on a Riemann surface such that every rational function is ramified at this point? Let $X$ be a compact connected Riemann surface, and let $S\subset X$ be a finite subset. 
Does there exist a morphism $f:X\to \mathbf{P}^1(\mathbf{C})$ which is unramified at the points of $S$?
I'm interested in the case where $X$ is of genus at least $2$. (The genus zero case is trivial: take $f$ to be the identity.)
The answer is trivial when $S$ is empty. (Any morphism $f:X\to \mathbf{P}^1(\mathbf{C})$ will do.)
Let $h:X\to \mathbf{P}^1(\mathbf{C})$ be a morphism with ramification locus $R(h)$. Then, if $S\subset X\backslash R(h)$, the answer is yes. 
How effective can our answer be? That is, suppose that there exists such an $f$. Then, can we bound its degree?
The title is a special case of the above question: take $S=\{\textrm{pt}\}$.
 A: If $S=\lbrace P\rbrace $, put an algebraic structure $X^{alg}$ on $X$  and take a uniformizing parameter $ t\in \mathcal O_{X^{alg},P}$ at $P$.
Since $\mathcal O_{X^{alg},P}\subset Rat(X^{alg})=\mathcal Mer(X)$, the meromorphic function $t$ seen as a map   $X\to \mathbb P^1(\mathbb C)$ solves your problem.
You can generalize that to the case where $S$ is an arbitrary finite set, again by putting an algebraic structure on the Riemann surface $X$: look up  Corollary 1.16 in   Chapter VI of Miranda's  book, which solves your problem.
Edit: Let me however give a self-contained proof using Riemann-Roch.
Fix an arbitrary point $x_0\in X\setminus S$ outside of $S=\lbrace x_1,...,x_n \rbrace$.
Consider the divisors (where $N$ will be determined later)
$$D_1=(-2)\cdot x_1+...+(-2)\cdot x_n+N\cdot x_0\quad \text {and} \quad D_2=(-1)\cdot x_1+...+(-1)\cdot x_n+N\cdot x_0$$
and their associated sheaves (=line bundles) $\mathcal O(D_1), \mathcal O(D_2)$.
They give rise to a short exact sequence of sheaves
$$ 0\to \mathcal O(D_1)\to  \mathcal O(D_2)\to  \mathcal Q\to 0        $$  where the quotient sheaf $\mathcal Q$ is a finite sum of sky-scraper sheaves of rank $1$.
Taking cohomology we get $$ ...\to H^0(X, D_2)\to H^0(X,\mathcal Q)         \to H^1(X,D_1)\to ...$$
Now $H^1(X,D_1)$ is dual to $H^0(X,K_X(2\cdot x_1+...+2\cdot x_n-N\cdot x_0))$  (by Serre duality)  and is thus zero for $N\gt 2n+2g-2 $.
This choice of $N$ implies that the morphism $\gamma : H^0(X, D_2)\to H^0(X,\mathcal Q)$ is surjective.
And what has this got to do with your problem? It solves it!
Indeed, if you choose a coordinate $z_i$ near $x_i$, the stalk  $\mathcal Q_{x_i}$ is identified with the complex line $\mathbb C\cdot z_i$ and by choosing a  section $s\in H^0(X, \mathcal O(D_2))$ mapping to a $\gamma(s)\in H^0(X,\mathcal Q)$ non-zero at every $x_i$ you obtain the required meromorphic function $s$: its only pole is at $x_0$ and it  has zeros of order exactly $1$ at the $x_i$'s.
A: You can always get an $f$ of degree $\max(g+1,n)$, where $g$ is the genus and $n$ the number of points in $S$. I don't think it is a good estimate (the problem is that I construct $f$ so that every point of $S$ is a (simple) pole; there should be $f$'s with lower degrees if the values at points of $S$ are different)
Here is how to see it. Let $P_1,\dots,P_n$ be the points in $S$ and let $z_i$ be a local coordinate around $P_i$ (s.t. $z_i(P_i)=0$) on some disc $D_i\subset \Sigma$. Suppose the discs don't overlap. The function $1/z_i$ on $D_i\cap (\Sigma -P_i)$ gives a class $\alpha_i\in H^1(\Sigma,\mathcal{O})\cong\Omega^1(\Sigma)^*$ (use $D_i$ and $\Sigma-P_i$ as an open cover of $\Sigma$; $1/z_i$ is a function on $D_i\cap(\Sigma-P-i)$). We know that $\alpha_i\neq0$ (if $\alpha_i$ were a coboundary then $1/z_i=h-k$, where $h$ is holomorphic on $\Sigma-P_i$ and $k$ on $D_i$, but that means that $h$ is meromorphic on $\Sigma$ with a unique pole at $P_i$, which is impossible if $g>0$). 
We can suppose $n>g$ (if not then add some more points to $S$). As $\dim H^1(\Sigma,\mathcal{O})=g$ and all $\alpha_i$'s are non-$0$, there are $c_1,\dots,c_n\in\mathbb{C}$, all non-$0$, such that $\sum c_i\alpha_i= 0\in H^1(\Sigma,\mathcal{O})$. If you write $\sum c_i\alpha_i$ as a coboundary for the open cover $\Sigma-S,D_1,D_2,\dots,D_n$ of $\Sigma$, the holomorphic function on $\Sigma-S$ is a meromorphic function on $\Sigma$ with a simple pole at every point of $S$, hence non-ramified at $S$, and its degree is the number of poles (i.e. $n$).
edit I changed my answer completely as it contained a mortal gap
