= Solution
The same quotient $|x||y|^{-1/3}$ is unbounded near $(1,0)$, so the uniform Lipschitz hypothesis again fails. Two explicit solutions through the initial point are
$$
\boxed{y_1(x)=0,\qquad y_2(x)=\left(\frac{x^2-1}{6}\right)^3.}
$$
For the second, put $u=(x^2-1)/6$. Then $y_2'=3u^2(x/3)=xu^2=x(y_2)^{2/3}$, and $y_2(1)=0$. The real power here means the square of the real cube root, so the identity also holds when $y_2$ is negative. Both functions are differentiable through their zeros. Thus uniqueness fails in this case as well.
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