Solution: Note that the function $f(y) = y^{\frac{1}{3}}$ is not Lipschitz continuous on any interval containing the origin and hence the solution need not be unique. To see it is not Lipschitz, note that \begin{align*} \frac{f(y)-f(0)}{y-0} = \frac{y^{\frac{1}{3}}}{y} = \frac{1}{y^{\frac{2}{3}}}. \end{align*} Since the above difference quotients is not bounded on any interval containing $x=0$, the function $f(y)=y^{\frac{1}{3}}$ is not Lipschitz continuous on any interval containing $0$. Note that if $f:\mathbb{R} \to \mathbb{R} $ is a differentiable function, then it is Lipschitz if and only if the derivative is bounded.

Observe that one solution can be obtained by just integrating the differential equation. \begin{align*} \frac{2}{3} \frac{\mathrm{d}y }{\mathrm{d} x} = y^{\frac{1}{3}} & \implies \frac{2}{3} \frac{\mathrm{d} y}{y^{\frac{1}{3}}} = \mathrm{d} x \\ & \implies \int \frac{2}{3} y^{-\frac{1}{3}} \mathrm{d} y = \int \mathrm{d} x \\ & \implies \frac{2}{3} \left( \frac{y^{-\frac{1}{3} + 1}}{-\frac{1}{3}+1} \right) = x + c \\[1ex] & \implies \frac{2}{3}\left( \frac{y^{\frac{2}{3}}}{\frac{2}{3}} \right) = x+ c \\[1ex] & \implies y^{\frac{2}{3}} = x + c \\[1ex] & \implies y(x) = (x+c)^{\frac{3}{2}}. \end{align*} The initial condition implies $c=0$ and therefore, one solution of this differential equation will be $y(x) = x^{\frac{3}{2}}$.

Another solution is $y\equiv 0$.

Now we will show that this differential equation has infinitely many solutions. For that let $a>0$ be any real number. Let \[ y(x) = \begin{dcases} 0, &\text{ if } x\leq a ;\\ (x-a)^{\frac{3}{2}}, &\text{ if } x > a. \end{dcases} \] Note that the above is also a solution and hence, we conclude that the given differential equation has infinitely many solutions. The above solutions are demonstrated in the below figure.