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.
As mentioned by Cantlog there is true if $X$ is integral and separated. Here is a counterexample showing this is not true in the integral non-separated case.
Take $X$ to be the affine line with doubled origin, $x$ the north pole and $y$ the south pole. Let $f(t) = p(t)/q(t) \in k(t)$ be any rational function on $X$. We see $$f(x) = f(y) = \text{constant coefficient of $p(t)$/ constant coefficient of $q(t)$}$$
but $x \neq y$.