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Is there a neighborhood of $(0,0,0)$ on $V(x^2-y^3, y^2-z^3)$ that is isomorphic to an open subvariety of a plane curve?

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No, that cannot possibly happen. First recall that for every plane curve $C \subset \Bbb{P}^2_k$, and every point $Q \in C$, the cotangent space $\mathfrak{m}_Q/\mathfrak{m}^2_Q$ is a $k$-vector space of dimension at most two.

On the other hand, you can show that for $P=(0,0,0)$ the cotangent space $\mathfrak{m}_P/\mathfrak{m}^2_P$ of $V(x^2-y^3,y^2-z^3)$ at $P$ is three-dimensional. Indeed, $\mathfrak{m}_P/\mathfrak{m}^2_P$ is generated by the residues $$ x,y,z \mod \mathfrak{m}^2_P $$ and you can check that these elements are linearly independent over $k$ (which in turn essentially boils down to the fact that the ideal $(x^2-y^3,y^2-z^3)$ contains no linear polynomials).

Therefore, the cotangent space of $V(x^2-y^3,y^2-z^3)$ at $P=(0,0,0)$ cannot be isomorphic to the cotangent space of a plane curve, and hence an isomorphism as described in your question cannot exist.

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