Do differentiable functions preserve measure zero sets? Measurable sets?

If $f$ preserves null-sets, it also preserves Lebesgue-measurability (but not necessarily Borel measurability), because by regularity, every Lebesgue measurable set $M$ can be written as

$$ M = N \cup \bigcup K_n $$

with $K_n$ compact and $N$ a null-set. By continuity, $f$ preserves compact sets.

Now a theorem in Rudin, Real and Complex Analysis (Lemma 7.25) shows in particular that every everywhere differentiable function maps null-sets to null-sets, so that your claim is true.

The following claim implies your result.

Claim: Let $E \subseteq \mathbb{R}$ be arbitrary. Suppose $|f'(x)| \leq M$ for all $x \in E$. Then $\mu^{\star}(f[E]) \leq M\mu^{\star}(E)$.

Proof: Fix $\epsilon > 0$. Get an open set $U \supseteq E$ with $\mu(U) < \mu^{\star}(E) + \epsilon$. For each $x \in E$, let $\delta_x > 0$ be such that $(x - \delta_x, x + \delta_x) \subseteq U$ and $|f(y) - f(x)| \leq (M + \epsilon)|y - x|$ for every $y \in (x - \delta_x, x+ \delta_x)$. Consider the family of closed intervals: $V = \{[f(x), f(y)], [f(y), f(x)] : x \in E, |y - x| < \delta_x\}$ - This is a Vitali covering of $f[E]$ which means that every point in $f[E]$ is covered by arbitrarily small intervals in $V$. Using Vitali covering theorem, get a countable subfamily $C$ of pairwise disjoint intervals which covers all but a null part of $f[E]$. The (disjoint) union of the intervals in $C$ has measure at most $(M + \epsilon) \mu(U)$ which is less than $(M + \epsilon)(\mu^{\star}(E) + \epsilon)$. Now, let $\epsilon$ go to zero.