An ordinary differential equation with non Lipschitz RHS 
I have the following initial value problem: 
  $$\tag{IVP}\label{IVP}
\begin{cases}
x^\prime(t) = -\sqrt[3]{x}\\
x(0) = x_0
\end{cases}
$$
  and I have to show that for every $x_0 \in \mathbb R$ a solution to \eqref{IVP} exists and it is unique. 

Existence is not an issue, as the function $f(x) := -\sqrt[3]{x}$ is continuous, hence Peano's Theorem applies. But what about uniqueness? 
Here are my considerations. As long as the solution is positive/negative it is decreasing/increasing. Let us suppose $x_0<0$. Then for $t$ small enough it remains negative so that I can divide 
$$
\frac{x^\prime(t)}{\sqrt[3]{x}} = -1
$$
which by integration yields
$$
\int_{x_0}^{x(t)} \frac{dy}{\sqrt[3]{y}} = - t, \qquad \text{i.e. } x(t) = \left(-\frac{2}{3}t + x_0^{2/3}\right)^{3/2}
$$
and here comes already a problem: the formula I have found implies $x\ge 0$ (it is a square root!). Where am I mistaken? How can I show uniqueness? Thanks. 
 A: For $x_0 = 0$ there is multiplicity of solutions.
Indeed, one solution is $x(t) \equiv 0$, and another one is
$$
y(t) :=
\begin{cases}
(-\frac{2}{3} t)^{3/2}, & \text{if}\ t < 0,\\
0, & \text{if}\ t \geq 0.
\end{cases}
$$
A: The problem is that the definition of $\sqrt[3]{\cdot}$ is ambiguous, if $x>0$ then there is no problem, i.e this is $x^\frac{1}{3}$, but for $x<0$ you can not use such formula so I suppose the definition is:
$$\sqrt[3]{x}=sgn(x) |x|^\frac{1}{3}$$
in particular the antiderivative for $x <0$ is not $\frac{3}{2} x^\frac{2}{3}$ but $sgn(x) \frac{3}{2} x^\frac{2}{3}$.

To be more precise as long as $x(t)<0$ with $z(t)=-x(t)>0$ we have:
$$-z'(t)=+z(t)^\frac{1}{3}$$
so:
$$z(t)=\left(z_0^\frac{2}{3}-\frac{2}{3}t \right)^\frac{3}{2}$$
i.e in terms of $x$:
$$x(t)=-\left((-x_0)^\frac{2}{3}-\frac{2}{3}t \right)^\frac{3}{2}$$
which is indeed negative and well defined fot $t<\frac{3}{2} (-x_0)^\frac{2}{3}$.

For uniqueness, the only case remaining is when $x$ touches $0$ (or $x_0=0$). But with $n(t)=x(t)^2 \geq 0$ you have:
$$n'(t)=2 x(t) x'(t) = - 2 |x|^\frac{4}{3} \leq 0$$
so $n$ is non increasing. In particular if $x(t^*)=0$ then for all $t \geq t^*$ we have $n(t)=0$ i.e $x(t)=0$.
