Show that the derivatives of a $C^1$ function vanish a.e. on the inverse image of a null set

The statement is true, below is a proof. You will see from the proof that it works under the assumption weaker than $C^1$, namely, it suffices to assume that $f$ is absolutely continuous on almost every line parallel to a coordinate axis.

I will first give a proof in the 1-dimensional case and then use it to proof the general claim. I will use the notation $|Z|$ to denote the (1-dimensional) Lebesgue measure of subsets $Z\subset {\mathbb R}$. I will also denote by $B_r(z)\subset {\mathbb R}$ the closed ball of radius $r$ centered at $z\in {\mathbb R}$. If $z=0$, set $B_r=[-r,r]$.

We will need

Theorem (Lebesgue density theorem). Let $Z\subset {\mathbb R}$ be a (measurable) subset of positive measure. Then almost every point $z\in Z$ is a density point, i.e., $$ \lim_{r\to 0} \frac{|B_r(z)\setminus Z|}{|B_r(z)|}=0. $$

Let $A\subset {\mathbb R}$ be a measure zero subset. For a $C^1$-smooth function $f: {\mathbb R}^n\to {\mathbb R}$ set $$ E=f^{-1}(A), E'= E\cap \{x: df(x)\ne 0\}. $$ Note that both $E$ and $E'$ are measurable sets.

Lemma. Let $f: {\mathbb R}\to {\mathbb R}$ is a $C^1$-function. Then $E'$ has zero measure.

Proof. Suppose not. Then there exists a density point $z\in E'$. By making affine change of variables in domain and range, we can assume that $z=0$ and $f'(0)=1$. By continuity of $f'$, there exists $r_0>0$ such that for all $x\in B_{r_0}$, $$ \frac{1}{2} < f'(x)< 2. $$ In particular, for every $r\in (0, r_0)$ and every measurable subset $S\subset B_r$, $$ |f(S)|\le 2|S| $$ (here is the only place where the assumption that the domain of $f$ is 1-dimensional is needed). Set $$ E_r'= B_r\cap E', E_r''= B_r \setminus E'. $$ Since $f(E_r')\subset A$, and $|A|=0$, we get $|f(E_r')|=0$. On the other hand, since $0$ is a density point of $E'$, for every $\epsilon>0$ there exists $0<r_1<r_0$ so that for all $r\in (0, r_1)$ $$ |E_r''|< \epsilon |B_r|. $$ Thus, $$ |f(E_r'')|\le 2\epsilon |B_r|. $$ By compining this with the fact that $|f(E_r')|=0$, we see that, for $r<r_1$, $$ |f(B_r)| \le 2\epsilon |B_r|. $$ On the other hand, by the mean value theorem (taking into account that $0.5\le f'(x)<2$ for $x\in B_{r}$, $$ |f(x)-f(0)|\ge \frac{1}{2}|x| $$ for all $x\in B_r$. In particular, $|f(B_r)|\ge |B_r|/2$ for all sufficiently small $r>0$. Thus, we obtain: $$ \frac{1}{2}|B_r|\le |f(B_r)|\le 2\epsilon|B_r| $$ for all $r\in (0,r_1)$. Taking $\epsilon=1/8$ we obtain a contradiction. qed

Now, we can prove the claim for general $n$.

Theorem. Let $u: {\mathbb R}^n\to {\mathbb R}$ is a $C^1$-function. Then the set $$ E'=u^{-1}(A)\cap \{p: du(p)\ne 0\}, $$ has zero measure.

Proof. It suffices to show that for each $i=1,...,n$, the set $$ E_i'=E\cap \{p\in {\mathbb R}^n: \frac{\partial}{\partial x_i} u(p)\ne 0\} $$ has zero measure. By renumbering the cooridnates, it suffices to check this for $i=n$. For every $y\in {\mathbb R}^{n-1}$ consider the line $L_y=\{(y,x): x\in {\mathbb R}\}$ in ${\mathbb R}^n$. Restriction of $f$ to $L_y$ is $C^1$-smooth. Therefore, by Lemma, for each $y$, the set $$ E_y'= E\cap \{\frac{\partial}{\partial x} f (y,x)\ne 0\} $$ has zero measure 1-dimensional. Therefore, by Fubini's theorem, the set $$ E_n'= \bigcup_{y\in {\mathbb R}^{n-1}} E_y' $$ has zero $n$-dimensional measure. qed