I want to find the local rings of $V(y^2-x^3)$, and establish if it's isomorphic to $K[x]_{(x)}$, or maybe some other ring which I don't know. We take $p=(t^2,t^3) \in V$ and we want to find $O_{(t^2,t^3)}(V) \cong \Gamma(V)_{m_p}$ where $m_p$ corresponds to the maximal ideal of $p$ in $\Gamma(V)\cong K[x,y]/(y^2-x^3) \cong K[X^2,X^3]$ where $X$ denotes an indeterminate. Since $I(\{p\})=(x-t^2,y-t^3)$ I get that $m_p=(\overline{x}-t^2,\overline{y}-t^3)$ which is sent to $(X^2-t^2,X^3-t^3)$ in $K[X^2,X^3]$ so we consider $K[X^2,X^3]_{(X^2-t^2,X^3-t^3)}$ and from now I'm a bit lost. Is what I did so far correct ?
At first I wanted to say that $\Gamma(V)_{m_p}\cong K[X^2,X^3]_{(X^2,X^3)}$, I thought that $m_p$ was sent to $(X^2,X^3)$ but with what I did above I'm not sure of it.
Is it possible to proceed with what I did ? Thank you
Yes, what you've done so far is correct.
If $t\neq 0$, then $X=\frac{X^3}{X^2}\in K[X^2,X^3]_{(X^2-t^2,X^3-t^3)}$ and $(X^3-t^3)-X(X^2-t^2)=t^2(X-t)$, so your local ring becomes $K[X]_{(X-t)}$. This is isomorphic to $K[X]_{(X)}$ by the map $X\mapsto X+t$.
If $t=0$, there's not much you can do to simplify $K[X^2,X^3]_{(X^2,X^3)}$.