It is important to know that finitely generated $k$-algebras are not just of the form $k[x_1,...,x_n]$. They can also be quotients of these rings by ideals. For instance $A=\mathbb{C}[x,y]/(x^2-y^3)$. This is a $\mathbb{C}$-algebra, but will not be isomorphic to any polynomial ring over $\mathbb{C}$. One way to see this is that its localization at the maximal ideal $(x,y)$ will not be integrally closed in the fraction field of $A$. The element $x/y$, for instance, will be integral over $A_{(x,y)}$ but not actually in $A$.
However, if we localize any polynomial ring $P$ over $\mathbb{C}$ at a maximal ideal $m$, we find that $P_m$ is integrally closed in the fraction field of $P$.
This may not be the best way to see that $A$ is not a polynomial ring, but it is very geometric. The ring $A$ is the coordinate ring of the plane curve given by $x^2 = y^3$. This curve has a cusp at the origin and is quite singular. This geometric fact is somehow wrapped up in the algebra of the local ring $A_{(x,y)}$. Any polynomial ring is the coordinate ring of a certain affine space. Affine spaces are nice and smooth, and this fact is again reflected somehow in the algebra of its localizations at maximal ideals.
What normalization tells us is that finitely $k$-algebras are reasonable objects. They are integral extensions (very nice extensions) of polynomial rings (very nice rings).
One geometric consequence (at least over an algebraically closed field) that normalization gives us is that any algebraic set is a cover of an affine space. Normalization gives us some inclusion of a polynomial ring $P=k[y_1,...,y_d]$ into the ring $A$. If we think of this geometrically, $A$ is the coordinate ring of some variety, and $P=k[y_1,...,y_d]$ is the coordinate ring of some $d$-dimensional linear subspace. The geometric picture is the reversal of the algebraic picture, in that an inclusion of $P$ into $A$ means the variety associated with $A$ projects (surjectively) onto the linear space associated with $P$. An important consequence of this is the Nullstellensatz.
I figured out where my mistake is. (I answer my own question in case someone run into the same problem.)
$\ker \varphi \neq\lbrace 0 \rbrace$ is okay. But to continue I need that $a_3$ be an algebraic integer over $k[a_1,a_2]$, so I need a monoic polynomial $g\in k[a_1,a_2][X]$ but that isn't always exist. What is true is that if $g\in\ker \varphi$, there exist suitable $\alpha ,\beta\in k$ such that $g(X+\alpha Z,Y+\beta Z, Z)$ is monoic in $Z$. In this example $\alpha =1$ and $\beta =0$ suit.
So $a_3$ is an algebraic integer over $k[a_1-a_3,a_2]$, and only now I consider
$$ \varphi ' : k[X,Y] \longrightarrow k[a_1-a_3,a_2]$$
One can prove that $\ker \varphi ' =\lbrace 0 \rbrace $, hence using Noether's normalization lemma we get that $k[a_1,a_2,a_3]$ is a finite $k[a_1-a_3,a_2]$-module and $a_1-a_3,a_2$ are algebracally independent over $k$.
Best Answer
I would map $A$ to $\Bbb C(X,Y)$ (the field of fractions of $\Bbb C[X,Y]$) via $x\to X$, $y\to Y$ and $z\to-xy/(x+y)$. The image of $A$ contains $X$ and $Y$ which are algebraically independent over $\Bbb C$, therefore so are $x$ and $y$ in $A$.