Localization of $R = k[X,Y]/(Y^2 - X^2 + X)$ is a DVR 
Let $R = k[X,Y]/(Y^2 - X^2 + X)$ where $k$ is a field (say of characteristic different from 2 and 3) and $m = (X,Y)$ an ideal. Show that the localization of $R$ at $m$, $R_m$ is a discrete valuation ring.

I can show that $R_m$ is a local Noetherian ring of dimension 1. So it also suffices to show that $R_m$ is a PID (equivalently, the unique maximal ideal is principal), normal, or regular.
Any ideas on how to proceed from here is greatly appreciated.
 A: In the local ring $R_m = k[X,Y]_m/(Y^2-X(X-1))$ the element $X-1$ is a unit, since it does not belong to $m$. Thus, $R_m = k[X,Y]_m/(uY^2-X)$ for a unit $u\in R_m$, which implies $R_m\cong k[Y]_{(Y)}$.
A: $\def\spec{\text{Spec}}$ $\def\ht{\text{ht}}$ Alternatively, since $m$ is a $k$-rational point $\spec(R)$, we know that the tangent space of $\spec(R)$ at $m$ is given by the kernel of the map $k^2\to k$ given by $\begin{pmatrix}1 & 0\end{pmatrix}$. But, clearly the kernel of this map is dimension $1$, and thus $\dim_k T_{\spec(R),m}=1$.
Now, it's obvious that $R$ has dimension $1$. Indeed, we know that 
$$\dim(R)=\dim k[X,Y]-\ht((Y^2-X^2+X))=2-1=1$$
and, since $\spec(R)$ is an integral variety we know that $R$ is catenary, and thus $\dim(R_m)=\dim(R)=1$.
Thus, $R_m$ is a regular local ring of dimension $1$ and thus a DVR.
A: The maximal ideal ${\frak m} = (X,Y)$ corresponds to the point $(0,0)$ on the curve $V(Y^2 - X^2 + X)$. The localization $R_{\frak m}$ is regular if and only if $(0,0)$ is a non-singular point of this curve. This last fact is easy to check: the partial derivates of $Y^2 - X^2 + X$ are $-2X + 1$ and $2Y$ and they do not both vanish at $(0,0)$.
