 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 areFor the second, put $u=(x^2-1)/6$. Then $y_2^{\prime }=3u^2(x/3)=x{u}^2=x(y_2{)}^{2/3}$, 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.