Cusp is homeomorphic to $\mathbb{R}^{n=1}$

Question: Show that the Cusp is homeomorphic to $\mathbb{R}^{n=1}$ and thus is a topological manifold.

The ordinary Cusp is given by $f\left ( x,y \right )=x^{2}-y^{3}=0$

In 2D, $f\left ( x,0 \right )=x^{2}$ Let the Cusp C and $\mathbb{R}$ be topological space.

Define:

$f: C\rightarrow \mathbb{R}$

$x \mapsto x^2$

Clearly, f is a bijective map.

To show that f is homeomorphic, it remains to to show that f and $f^{-1}$ is continuous. It remains to be shown that $f$ and $f^{-1}$ maps open sets between the given topological spaces.

Recall: $\forall n \in \mathbb{Z}^{+}, \mathbb{R}^{n}$ is a topological manifold

So clearly, here $\mathbb{R}^{n=1}$ is locally euclidean and there must exists a function f that maps open sets from $\mathbb{R}^{n=1}$ to the Cusp. Assume otherwise and arrive at a contradiction.

Or

Recall: If M is a topological space and U is an open subset of M then U is a topological manifold.

The second theorem seems more direct. $C \subseteq \mathbb{R}^{n=1}$ and so is a subspace of $\mathbb{R}^{n=1}$. By definition, the basis for the subspace topology on C is the collection of an arbitrary union of open sets in C and so C is an open set in $\mathbb{R}^{n=1}$. Therefore, C is a topological manifold and so C is a homeomorphism.

Have I blundered anywhere?

A possible homeomeorphism $f:\mathbb R\to C$ is given by $f(t)=(t^3,t^2)$.
$$f^{-1}:C\to \mathbb R:\left\{ \begin{array}{l} (x,y) \mapsto\frac xy \;\operatorname{for} \;(x,y)\neq (0,0) \\ (0,0)\mapsto 0 \\ \end{array}\right.$$