Problem: Find all the solutions of the IVP $$x'=3x^{\frac{2}{3}}, x(0)=0$$ for $t\geq 0$.
Here $3x^{\frac{2}{3}}$ is not $C^1$, so the existence and uniqueness theorem does not apply here.
My guess the solutions is $$x=\left\{\begin{matrix} 0 & \text{if }0\leq t< t_{0} \\ (t-t_{0})^3 & \text{ if }t\geq t_{0} \end{matrix}\right.$$ $t_{0}\in\mathbb{R^+}$ or $t_{0}\rightarrow+\infty$.
But my professor told me that there are a lot more!
My main question is: How can I find all the solutions, and then prove that they are all the solutions, rigorously?
First of all, we can build $x'$, so $x$ is continuous. Since $x^{2/3}=(x^{1/3})^2$, we have $x'\ge 0$, so $x$ is increasing. Then $x$ is either constant, zero, on $J:=[0,\infty)$, or there exist a maximal $t_0=\sup x^{-1}(\ \{0\}\ )=\max x^{-1}(\ \{0\}\ )$ in $J$ so that $x(t_0)=0$. We will only consider this last case, $t_0<\infty$.
Then we have $x'>0$ on $(t_0,\infty)$. Now we forget about the condition in $0$ for a while, solve the given differential equation in $(t_0,\infty)$, knowing $x>0$. On $(0,\infty)$, the function $t\to t^{1/3}$ is of class $C^1$, so we can build as compositum the differentiable function $$ y = x^{1/3}\ , $$ which satisfies by the chain rule $$ y'= (x^{1/3})' =\frac 13x^{-2/3}\cdot x' =\frac 13x^{-2/3}\cdot 3 x^{2/3}=1\ . $$ From $y'=1$ on $(t_0,\infty)$ we find a constant $C$ with $y(t)=t+C$ on $(t_0,\infty)$. So far we know about $x$ that it is $0$ on $[0,t_0]$, it is $t\to y^3(t)=(t+C)^3$, and it is continuous in $t_0$. The only matching constant is $C=-t_0$. We get the solutions and only the solutions from the OP. (It is clear that these are of class $C^1$ and satisfy the given differential equation.)