Proving that $x^2+2y^2+3z^3 = 1$ is an embedded manifold

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I am working on the following exercise:

Consider $S = \{ (x,y,z) \in \mathbb{R}^3 : x^2 + 2y^2 + 3z^3 = 1 \text{ and } z>0\}$

Show that $S$ can be parametrised as a graph of a function from an open set in $\mathbb{R}^2$ to $\mathbb{R}$. Further show that $S$ is an embedded manifold.

I made the following parametrisation:

Let $X = \left(-1,1\right) \times \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $g(x,y) = \frac{\sqrt{1-x^2+2y^2}}{3}$. Then we have our desired regular parametrisation $f$ for $S$ by:

$$f(x,y) = (x, 2y^2, g(x,y))$$

To show that $S$ is embedded I need to show that for every $x \in X$ and for all open $V \subseteq X$ there is an open $U \subseteq \mathbb{R}^n$ with $f(x) \in U$ such that $f(V) = U \cap S$.

I tried to do this using $\epsilon$-balls but I can not find some suiting bounds. Could you help me?