By definition the solution $x: \mathbb{R}\rightarrow \mathbb{R}$, $t \mapsto x(t)$ of a first order autonomous differential equation
$$\dot{x} := \frac{dx}{dt} = f(x)$$
for some arbitrary function $f(x)$ must obey
$$x(t) = x(t') \rightarrow \dot{x}(t) = \dot{x}(t')$$
Looking at a graph like this
one sees that $x(t)$ must be monotone to be the solution of a first order autonomous differential equation, i.e. there must be no two $t, t'$, $t < t'$ with $x(t) = x(t')$, because there would have to be a third $t''$ between $t$ and $t'$ with $x(t'') = x(t) = x(t')$ but $\dot{x}(t'') \neq \dot{x}(t) = \dot{x}(t')$.
My question is:
Let $x(t)$ be an arbitrary smooth and monotone function and $\dot{x}(t)$ its (smooth) derivative. Consider the set $f$ of ordered pairs
$$f = \{\langle x(t),\dot{x}(t)\rangle\ |\ t \in \mathbb{R}\}$$
which by definition is a function $f: \text{range}(x) \rightarrow \text{range}(\dot{x})$. Can it be shown that $f$ must be smooth?
(Maybe this is a standard exercise?) If not so: What are specific counter-examples?
The other way around: Since $x(t)$ must be monotone, $\dot{x}(t)$ has to be positive or negative (for all $t\in\mathbb{R}$).
Can it be shown that for each smooth positive or negative function $f(x)$ there is a smooth monotone function $x(t)$ that is the solution of $\dot{x} = f(x)$?
If not so: What are specific counter-examples?
The standard counter-example for a smooth inverse for a monotone function is $x(t)=t^3$ at $t=0$. As an autonomous ODE this gives $$ \dot x(t)=3t^2=3x(t)^{2/3} $$ which is not smooth at $x=0$.
If you can exclude that using generic terms you should be fine.
For the second question, consider the standard example $f(x)=x^2$. It does not have solutions over the entirety of $\Bbb R$. On the other hand, a standard exercise is to show that for linearly bounded $f$ you get global solutions.