I am trying to find the normalisation of the cuspidal curve $0=z^2 -w^3$ without its singularity, and following Simon Donaldson's book on Riemann surfaces, it seems I should find the critical points $R$ of the projection onto the first coordinate $\pi\colon X\backslash \{(0,0)\} \to S^2 \backslash \{0,\infty\}$ where $X$ is the curve and $S^2$ is the Riemann sphere. So to do this, I chose the following charts: for X, I chose the chart $\varphi(z,w)= \frac{z}w$ with inverse $\varphi^{-1}(z)=(z^3 ,z^2 )$, and for $S^2$ I chose the identity, $\psi\colon S^2\backslash\{\infty\}\to \mathbb C$. However, then the local map becomes $$ (\psi\circ \pi\circ \varphi^{-1})(z)=(\psi\circ \pi)(z^3,z^2)=\psi(z^3)=z^3 $$ which, according to Donaldson's definition* means that all the points are critical points! This cannot be the case, as the set should be discrete, so what did I do wrong?
*Donaldson's definition says that the critical points of $\pi$ is the points where the local map has the form $z^k$ with k>1, and that seems to every point.
I do not know why you want to project on the first coordinate. This seems to be not a very good thing in general, as the projection could be critical only because the curve is tangent to vertical lines, but maybe could be OK for this curve. The map $z^3$ however is not critical for $z\not=0$, this is then maybe the problem you have.
Anyway, the normalization of the curve is simply $\mathbb{C}$ and the normalization from $\mathbb{C}$ to your curve is exactly the map $\varphi^{-1}$ that you define, which sends $z$ onto $(z^2,z^3)$, and which is a bijective morphism. It induces moreover an isomorphism $\mathbb{C}\setminus \{0\}\to X\setminus \{(0,0)\}$ whose inverse is $\varphi$, that sends $(z,w)$ onto $\frac{w}{z}$. Hence, every point of $X\setminus \{(0,0)\}$ is non-critical (smooth). The point $(0,0)$ is however not smooth. You can see this by derivating the map $\varphi^{-1}$ at the origin, or by looking at the tangent space of $X$ at $(0,0)$.