\begin {align*} y^{\prime } & =y^{\frac {1}{3}}\\ y\left ( 0\right ) & =0 \end {align*}
First we find the region where solution exists and is unique. \(f\left ( x,y\right ) =y^{\frac {1}{3}}\). The domain of \(y^{\frac {1}{3}}\) is \(y\geq 0\) since we do not want complex values. Hence solution exists. The domain of \(f_{y}=\frac {1}{3}\frac {1}{y^{\frac {2}{3}}}\) is \(y>0\). Hence the region is all \(x\) and \(y>0\). i.e. the top half of the plane not including \(x\)-axis. Since the point given is \(\left ( 0,0\right ) \) on the \(x\)-axis, then the theory do not apply. There is no guarantee solution is unique. Only way to find out is to try to solve the ode and find out. Solving the ode gives \begin {align*} \int \frac {dy}{y^{\frac {1}{3}}} & =\int dx\\ \frac {3}{2}y^{\frac {2}{3}} & =x+C \end {align*}
Applying IC gives \(C=0\). Hence solution is\[ \frac {3}{2}y^{\frac {2}{3}}=x \] Solving for \(y\)\[ y^{2}=\left ( \frac {2}{3}x\right ) ^{3}\] Taking the square root of both sides gives\begin {align*} y & =\pm \sqrt {\left ( \frac {2}{3}x\right ) ^{3}}\\ & =\pm \left ( \frac {2}{3}x\right ) ^{\frac {3}{2}} \end {align*}
So there are two solutions. There is also a trivial solution \(y=0\). We see that the solution exists but not unique.