Your map $f:\mathbb C\to \mathbb C^2: w\mapsto (x,y)=(w^3,w^2)$ is a homeomorphism onto its image (endowed with the subspace topology inherited from $\mathbb C^2$).
Thus the image is a topological manifold since $\mathbb C$ is one.
Since $f$ is clearly continuous and bijective, the only non-trivial point in the argument above
is that $f$ is a closed map.
And the reason for that is that $f$ is proper ( proper maps are closed !): indeed, when $w$ tends to infinity $f(w)$ tends to infinty too.
[As an aside, the image of $f$ is the algebraic subset of $\mathbb C^2$ with equation $x^2-y^3=0$ ]