= Solution
For $x\ne0$, all component functions are continuously differentiable and the <Jacobian matrix> is
$$
\boxed{Df(x,y)=
\begin{pmatrix}
\dfrac1{3x^{2/3}}&2y\\[4pt]
0&5y^4
\end{pmatrix}}.
$$
At a point $(0,y_0)$, the first component cannot be <differentiable>: along the first coordinate axis its difference quotient contains $h^{1/3}/h=h^{-2/3}$, which is unbounded as $h\to0$. Therefore $f$ is <continuously differentiable function>[continuously differentiable] exactly at the points with $x\ne0$.
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