Let $X$ be an irreducible, normal variety over an algebraically closed field of characteristic zero. Let $x,y\in X$ be two points such that $f(x)=f(y)$ for every $f\in K(X)$ which is defined at $x$ and at $y$. Can I conclude that $x=y$?

I feel the answer should be affirmative. In fact, the statement can be reduced to the following: Given two effective prime divisors ${D}_{x}$ and ${D}_{y}$ on $X$, there exists a rational function $f\in K(X)$ with ${v}_{{D}_{x}}(f)\ne 0$ and ${v}_{{D}_{y}}(f)=0$.

If this is true, then assuming $x\ne y$ we could find a divisor ${D}_{x}$ containing $x$ but not $y$ and a divisor ${D}_{y}$ containing $y$ but not $x$, so a function $f$ as above would yield a contradiction. However, I just can't prove the statement, even though I also think it should be true.

I feel the answer should be affirmative. In fact, the statement can be reduced to the following: Given two effective prime divisors ${D}_{x}$ and ${D}_{y}$ on $X$, there exists a rational function $f\in K(X)$ with ${v}_{{D}_{x}}(f)\ne 0$ and ${v}_{{D}_{y}}(f)=0$.

If this is true, then assuming $x\ne y$ we could find a divisor ${D}_{x}$ containing $x$ but not $y$ and a divisor ${D}_{y}$ containing $y$ but not $x$, so a function $f$ as above would yield a contradiction. However, I just can't prove the statement, even though I also think it should be true.