Consider $Y = V(y^2-x^3) \subseteq \mathbb{A}^2$
Now, $\phi: \mathbb{A} \to Y, t \mapsto (t^2,t^3)$ is a birational map, but the pullback $\phi^\ast: K[x,y]/(y^2-x^3) \to K[t], x \mapsto t^2, y \mapsto t^3$, is not an isomorphism of K-algebras.
(Do we say: $\phi$ is not an "isomorphism of varieties"?)
Now if we localize in the origin we get an isomorphism of the local rings at $0$: $(K[x,y]/(y^2-x^3))_{(x,y)} \rightarrow K[t,t^{-1}]$, which gives us the "rational" parametrization $t \mapsto (t^2,t^3)$.
Is this correct?
Now I would like to handle the following more complicated example:
Consider the surface $X = V(y(x^2+z^2)-x) \subseteq \mathbb{A}^3$.
How to find here a local birational map from $X$ at $0$ onto $\mathbb{A}^2$ at $0$?
Of course we cannot simply take $\displaystyle (x,y,z) \mapsto \left(x,\frac{x}{x^2+z^2},z\right)$ because we get $0$-denominator in the origin.
What is the usual strategy to find the desired birational map?
How to find a good direction of projection?