Riemann surface $y=(1-x^3)^{1/3}$ Let $X\subset \mathbb C^2$ be the Riemann surface given by （the multivalued function) $y=(1-x^3)^{1/3}$, and let $\phi:X \to \mathbb C $ be the induced map. Let $X'\subset P(\mathbb C^2)$ be the complex curve $x^3+y^3=z^3$ (in homogeneous coordinates). I want to define a complex manifold structure on $X'$ and a holomorphic $\phi':X'\to \mathbb C$ that extends $\phi$.
So far I know the following,
(i) $(x,y) \mapsto [x:y:1]$ gives a homeomorphism $\psi$ of $\mathbb C^2$ onto its image.
(ii) $\psi(X)\bigcup \{[1:-\zeta_3:0], [1:-\zeta_3^2:0], [1:-1:0]\}=X'$.
So I want to first define a manifold structure for $X$ and push it to $X'$ by $\psi$ and use another chart to cover $\{[1:-\zeta_3:0], [1:-\zeta_3^2:0], [1:-1:0]\}$. I think $x$ is a coordinate chart around $x_0$ if we fix a branch (fix a $y_0$). So we need three charts to cover $X$ because we have three branches. Is this correct?
Also, how do I construct a compatible chart to cover the three points at infinity?

If I got well, $X$ is the zero locus $x^3+y^3=1$ in $\mathbb{C}^2$ and your $\phi $ is the projection on one of the coordinate (I will assume is the $y$ just to be clear).
Now you take $X'$ that is ,as you can easily prove, the projective closure of $X$ in $\mathbb{P}^2(\mathbb{C})$.
As you correctly said , you need three charts because intuitively there are three branches. A standard way to proceed in the projective spaces is to use the standard charts $x_i \neq 0$ that are (in the projective plane) biholomorphic to $\mathbb{C}^2$.
If you restrict $X'$ to this charts , you will get three varieties: the first one (in the chart $z \neq0$) is your $X$ (and so you get your immersion) and the other two are very closely related to $X$ (so you can easily find complex structures on them in the same way you did with $X$).
