What smooth functions are solutions of an autonomous ODE? Let $y$ be a smooth function, say $y : \mathbb{R} \rightarrow \mathbb{R}$. When can we find a continuous map $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $y'=f(y)$ ?
Obviously it's not always the case, since there are in general no reasons that $y'$ is constant on level sets of $y$; and it's not too difficult so see that it's possible when $y$ is monotonous. 
But is there a natural way to see this ? One that could possibly generalize to higher dimension in the range of $y$ ?
 A: 
it's not too difficult so see that it's possible when $y$ is monotone

And it's impossible when $y$ is not. Monotonicity of $y$ is equivalent to all sublevel sets $U_t=\{x: y(x)<t\}$ being connected (i.e., open intervals). Suppose there is a $t$ for which this set is not connected. Then it has at least two connected components, which are open intervals. Consequently, there is a point  $a$ on the boundary of $U_t$ such that $(a,a+\epsilon)\in U_t$ for some $\epsilon$. Take a sequence $x_n\searrow a$ such that $y'(x_n)<0$; such a sequence exists because $y$ is not constant to the right of $a$. Note that $y(x_n)\nearrow t$. 
Since $y'$ is negative whenever $y=y(x_n)$, the set $\{x: y(x)<y(x_n)\}$ is an interval of the form $(c,\infty)$. But $U_t$ is the union of these sets, hence it is also an interval. Contradiction. 
Higher-dimensional analog is a function $\vec r(t)$ taking values in $\mathbb R^n$. The necessary and sufficient condition for existence of continuous $f$  is for $\vec r'(t)$ to be a continuous function of the range of $\vec r$, and be uniformly continuous on bounded subsets of the range. The necessity is clear; sufficiency follows from Tietze extension.
