For the purposes of this question, assume that when "measure" is mentioned, we refer to the Lebesgue measure.
Suppose I have a function $f : \mathbb{R} \to \mathbb{R}$ such that $\frac{df}{dx}$ is non-zero at every point. Is it then the case that for any set of positive measure $X$, $f(X)$ also has positive measure? If not, what precondition is required for this to be the case.
I'm not sure if it makes a difference, but I would be satisfied with an answer for the case: $f : [a,b] \rightarrow \mathbb{R}$
Yes. By Darboux's theorem, $f'$ must have constant sign, so $f$ is monotone. It follows that $f$ is locally absolutely continuous (see Non-decreasing and everywhere differentiable on $[0,1]$ implies absolutelly continuous?), and so for any interval $I$, $$\mu(f(I))=\int_I |f'|.$$ Both sides of this equation are measures when considered as functions of $I$ (here we use the fact that $f$ is injective so $f(\cdot)$ preserves disjointness of sets), and it follows that the two sides are equal for all Borel sets as well: for any Borel set $X$, $$\mu(f(X))=\int_X|f'|.$$ If $X$ has positive measure, then the integral on the right is positive since $f'$ is always nonzero, so $\mu(f(X))>0$.