How to solve $\dot{x} = \frac{f(x)}{\|f(x)\|}$? How to solve the following ODE?
$$\dot{x} = \frac{f(x)}{\|f(x)\|},$$
where $x : \mathbb{R} \to \mathbb{R}^n$, i.e., $x(t)$ is the trajectory. The right-hand side $f : \mathbb{R}^n \to \mathbb{R}^n$ is a continuously differentiable function with respect to $x$. $\|\cdot\|$ is any vector norm. 
I think the right-hand side of the ODE is not Lipschitz continuous. For example, let's take $n=1$ and $f(x)=-x$, then the right-hand side $-x/|x|$ is not even continuous. In this case, the theorem of existence and uniqueness of a solution cannot be applied. Then how to analyze the existence of the solution? Is there finite escape time?
 A: A  qualitative approach (liberally assuming $f$ does not vanish when necessary) is to solve the equation up to a time change and then reparameterize (which, after writing what follows, I realize is exactly what Lutz Lehmann suggests doing in a comment). To be more specific, consider instead
$$\dfrac{dy}{dt}(t)=f(y(t)).$$
$f$ is $C^1$ (in particular locally Lipschitz), so the existence and uniqueness theorem guarantees that for any $y_0\in \mathbb{R}^n$, there is a unique $y\in C^1(\mathbb{R},\mathbb{R}^n)$ satisfying
$$\dfrac{dy}{dt}(t)=f(y(t))\text{ and } y(0)=y_0.$$
Let $x_0\in \mathbb{R}^n$ and let $y$ be the unique solution as above with $y(0)=x_0$. The arclength of $y$ is
$$s(t)=\int_0^t \left\vert\dfrac{dy}{dt}(\tau)\right\vert\, d\tau = \int_0^t \left\vert f(y(\tau))\right\vert\, d\tau.$$
If $f(x_0)\neq0$, since $f$ is continuous, for some time $T\in\mathbb{R}_{>0}$ we have that  $s:([0,T],0)\to(\mathbb{R},0)$ is strictly increasing (and is $C^2$), so that its inverse $t:([0,S],0)\to ([0,T],0)$ is well defined (and is $C^2$) (here $S=s(T)$). Take $[x:s\mapsto y\circ t(s)] \in C^1([0,S],\mathbb{R}^n)$. Then
$$\dfrac{dx}{ds}(s)=\dfrac{dy}{dt}(t(s))\,\dfrac{dt}{ds}(s)=\dfrac{f(y(t(s)))}{\dfrac{ds}{dt}(t(s))}=\dfrac{f(y(t(s)))}{\vert f(y(t(s)))\vert}=\dfrac{f(x(s))}{\vert f(x(s))\vert}$$

Rem.1: One can generalize this to arbitrary initial times with a little more care.
Rem.2: $f$ not vanishing at an initial position $x_0$ means that it does not vanish in a neighborhood of $x_0$, which means that one can similarly extend the solution in some finite negative time.
Rem.3: Note that the trajectories of the ODE with $y$ as the unknown partition $\mathbb{R}^n$ (this is not quite a foliation since we don't know if $f$ has zeros). Since the trajectories of both ODE's are the same up to a reparameterization, we have the phase portrait of the ODE with $x$ as the unknown.
