Let $W=V(y^{2}-x^{3}) \subseteq \mathbb{A}^{2}$ and $k$ algebraically closed. Clearly the dimension of the tangent space at the origin is $2$. I want to compute this using the definition the fact that $\operatorname{dim} T_{(0,0)}W=\operatorname{dim}_{k} \mathfrak{m}/\mathfrak{m}^{2}$. Where $\mathfrak{m}$ is the maximal ideal of the local ring $\mathcal{O}_{(0,0),W}$.
OK according to Hartshorne the local ring is isomorphic to the localization of the corresponding coordinate ring localized at the ideal $(x,y)$ right?
So we need to compute the maximal ideal of $k[x,y]/(y^{2}-x^{3})$ localized at $(x,y)$. I know that in general given a ring $A$ then if $\mathfrak{p} \in \operatorname{Spec}(A)$ we have $\mathfrak{p}A_{p}=\{\frac{g}{h}: g,h \in P\}$ is the unique maximal ideal. However I don't see how to simplify things here. How to compute this?