Let $n\in\mathbb N$ and $h:\mathbb R^n\to\mathbb R$ be partially differentiable with respect to the first variable. If necessary, assume that $h$ is (jointly) continuous$^1$.
If $n=1$, we can show that $|h|$ is differentiable at $x$ with $$|h|'(x)=\operatorname{sgn}(h(x))h'(x)\tag1$$ for all $x\in D_1:=\{h\ne0\}\cup\{h'=0\}$. It's easy to see that each point in $N_1:=\mathbb R\setminus D_1$ is an isolated point of $\{h=0\}$ and hence $N_1$ is countable. In particular, $N_1$ has Lebesgue measure $0$.
Does the conclusion remain to hold for general $n$, i.e. does the set $N_n\subseteq\mathbb R^n$ on which $|h|$ is not partially differentiable with respect to the first variable have Lebesgue measure $0$?
By the aforementioned result, we should still obtain that $|h|$ is partially differentiable with respect to the first variable at $x$ with $$\frac{\partial|h|}{\partial x_1}(x)=\operatorname{sgn}(h(x))\frac{\partial h}{\partial x_1}(x)\tag2$$ for all $x\in D_n:=\{h\ne0\}\cup\left\{\frac{\partial h}{\partial x_1}=0\right\}$. However, if $x\in N_n=\mathbb R^n\setminus D_n$, then we are not able to conclude that $x$ is an isolated point of $\{h\ne0\}$ anymore (are we?). The only thing which should follow now is that there is a $\delta>0$ with $$h(x+te_1)\ne0\;\;\;\text{for all }0<|t|<\delta\tag3.$$ In the case $n=1$, $(3)$ implies that $x$ is an isolated point of $\{h=0\}$.
$^1$ The particular $h$ I'm interested in is of the form $h(x)=\prod_{i=1}^nf(x_i)-1$, $x\in\mathbb R^n$, for some $f\in C^1(\mathbb R)$ with $f>0$ and $\int f(x)\:{\rm d}x=1$. So, feel free to add any assumption which is satisfied in this special case.