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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?

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Correct, $\phi$ is not an isomorphism of varieties: you can prove directly that it has no inverse by writing $K[x, y]/(y^2 - x^3)$ as $K[t^2, t^3]$. – Qiaochu Yuan May 23 '11 at 20:20

The answer to your question "Is this correct ?" after the sentence "Now if we localize in the origin we get an isomorphism of the local rings..." is no for two reasons:

I) The local ring of $\mathbb A^1_K$ at zero is not $K[t,t^{-1}]$, but $K[t]_{(t)}$, a ring in which you may invert all polynomials not multiple of $t$, whereas in $K[t,t^{-1}]$ it is almost the opposite: you can only invert the powers of $t$!
Anyway, $K[t,t^{-1}]$ has infinitely many maximal ideals [even for a finite field $K$], and is thus not local.

II) Even taking I) into account, $\phi ^\ast$ does not give an isomorphism of the local rings.
Actually the local rings $K[t]_{(t)}$ and $K[x,y]/(y^2-x^3))_{(x,y)}$ are not isomorphic at all since (for example) $K[t]_{(t)}$ is a PID whereas $K[x,y]/(y^2-x^3))_{(x,y)}$ is not a UFD.

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To answer part of your question, a birational map (or, a little bit more precisely, morphism) $\phi: X \rightarrow \mathbb{A}^2$ is simply given by the formula $(x,y,z) \mapsto (x,z)$.

Its inverse (not a morphism!) is given by something like your formula above: it is $\phi^{-1}: (x,z) \mapsto (x,\frac{x}{x^2+z^2},z)$. It's not hard to check that $\phi \circ \phi^{-1}$ and $\phi^{-1} \circ \phi$ are the identity functions on their respective domains of definition.

As you rightly say, this isn't defined at the point $(0,0)$, but that's ok: by definition a birational map from a variety only needs to be defined on a dense open subset of the variety. Here $\phi^{-1}$ is clearly defined on the subset $\mathbb{A}^2 \setminus \left\{ x = \pm \sqrt{-1} \ z \right\}$, which is dense in $\mathbb{A}^2$.

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