Let $f : \mathbb{R} \to \mathbb{R}$ be measurable and let $Z = {\{x : f'(x)=0}\}$. Prove that $λ(f(Z)) = 0$. The following is an exercise from Bruckner's Real Analysis:

Let $f : \mathbb{R} \to \mathbb{R}$ be measurable and let $Z = {\{x : f'(x)=0}\}$. Prove  that $λ(f(Z)) = 0$.

For the case $f$ being nondecreasing / nonincreasing, we can defined $g=f^{-1}$ and then $g'$ and then use the following theorem from the book :

Let $f$ be nondecreasing / nonincreasing / of bounded variation on $[a,b]$. Then $f$ has a finite derivative almost everywhere.

Is it possible to "reduce" evaluation of any measurable $f$ to a nondecreasing one or otherwise how the claim can be proved for any measurable $f$?

A slightly simpler way to proof:
We have that $x\in Z$ if and only if, for all $m$, there is $n$ , such that $\forall y, |x-y| \leq \frac{1}{n}  \implies |f(x)-f(y)| \leq \frac{1}{m} |x-y|$.
Let $Z_{m,n}= \{ x \in \Bbb R: \forall y, |x-y| \leq \frac{1}{n}  \implies |f(x)-f(y)| \leq \frac{1}{m} |x-y|\}$.
So we have $Z= \bigcap_m \bigcup_n Z_{m,n}$.
Now, for all $p \in \Bbb Z$ and $k \in \{ 0, ... , n-1\}$, we have
$$ \lambda^* \left(f \left( Z_{m,n} \cap \left [p+\frac{k}{n},p+\frac{k+1}{n} \right ]  \right ) \right ) \leq \frac{1}{m}\frac{1}{n} $$
where $\lambda^*$ is the Lebesgue outer measure.
Since
$$f \left( Z_{m,n} \cap \left [p,p+1\right ]   \right ) = \bigcup_{k=0}^{n-1}f \left( Z_{m,n} \cap \left [p+\frac{k}{n},p+\frac{k+1}{n} \right ]   \right )$$
we have, by sub-additivity,
$$ \lambda^* \left(f \left( Z_{m,n} \cap \left [p,p+1\right ]  \right ) \right ) \leq \frac{1}{m} $$
Now, note that, for all $m$ and $n$,  $Z_{m,n} \subseteq Z_{m,n+1}$, so $f(Z_{m,n} \cap [p, p+1]) \subseteq  f(Z_{m,n+1} \cap [p, p+1]) $. So (see Remark 1),
$$ \lambda^* \left(f \left( \bigcup_n Z_{m,n} \cap \left [p,p+1\right ]  \right ) \right )= \lambda^* \left(\bigcup_n f \left(  Z_{m,n} \cap \left [p,p+1\right ]  \right ) \right ) = \\ = \lim_{n \to \infty} \lambda^* \left(f \left( Z_{m,n} \cap \left [p,p+1\right ]  \right ) \right ) \leq \frac{1}{m} $$
Since $Z= \bigcap_m \bigcup_n Z_{m,n}$, we have, for all $p \in \Bbb Z$, $\lambda^* (f(Z \cap [p, p+1])) \leq \lambda^* (f(\bigcup_n Z_{m,n} \cap [p, p+1])) \leq \frac{1}{m}$, for all $m$.
So, for all $p \in \Bbb Z$,
$\lambda^* (f(Z \cap [p, p+1])) =0$. It follows that $\lambda^* (f(Z)) =0$. So $f(Z)$ is measurable and $\lambda(f(Z))=0$.
Remark 1: Here, we use the result:

if $ E_{n}  \subseteq  E_{n+1} $ then $\lim\limits_{n\mapsto \infty} \mu^*(E_n) = \mu^*(\bigcup_{n=1}^\infty E_n) $ even if each $E_n$ is a non-measurable set, where $\mu^*$ is Lebesgue outer measure (or any regular outer measure).

This can be easily proved using measurable covers and it is an exercise in Halmos (section 12 exercise 4).
Remark 2: If you know Lemma 7.10 in Bruckner's Real Analysis: you can use exercise 7.3.1 in the same book to easily prove that $Z = {\{x : f'(x)=0}\}$  is measurable and then the answer to your question becomes a direct trivial  application of Lemma 7.10. (However the proof itself of Lemma 7.10 depends on the proof of Lemma 7.9 and exercise 7.3.1 and so it is more complex than what is needed to directly answer the question as posted).
