Given a plane curve $C$ over an algebraically closed field of characteristic 0 and $p\in C$ a singular point, if $f:\tilde{C}\to C$ is a resolution of the singularity and $f^{-1}(p)$ consists of one point, is it true that locally the singularity is of the form $x^m=y^n$ with $m$ and $n$ coprime? Can anything be said if $f^{-1}(p)$ consists of two points?
I've been searching through the vast literature of singularities of plane curves but have been unable to find any good answer.
For any curve, not just planar and not just in characteristic 0, the resolution is just the normalization. Looking at the completed local ring $\widehat{\mathcal{O}}_{C,p}$, the normalization is isomorphic to $$\prod_{i= 1}^s k[[x_i]]$$ by the Cohen structure theorem for some $s$. Then the number of points in $f^{-1}(p)$ is exactly the $s$ in this decomposition. In particular, one can produce examples where $f^{-1}(p)$ is a single point by looking at finite index subrings of $k[[x]]$. These are planar if and only if they are generated as a $k$-algebra by $2$ elements.
For an example that is not locally isomorphic to $y^m = x^n$, not even at the level of the completed local ring, consider the subring $k[[t^4, t^6 + t^7]] \subset k[[t]]$ which is the completed local ring of the curve $y^4 = x^7 - x^6 + 4x^5y + 2x^3y^2$.