This is a homework question. I am looking for a hint on where to start with solving this question!
Let $U$ be an open subset of $\mathbb{R}^{2}$ and $K\subseteq U$ be compact. Suppose that $f:U\to \mathbb{R}$ is a continuously differentiable function. Let $E=\{(x,y)\in K: f(x,y)=0\}$. Suppose $df$ is never $0$ on $E$, that is, the Jacobian is non-zero on $E$. We are to show then that $E$ has volume (area) $0$ in $\mathbb{R}^{2}$.
I assume that we are to show that the Jordan volume of $E$ is $0$. Do we do so by showing that the Lebesgue measure of $E$ is $0$ or is there a direct way of proving this? In particular, where do we use the fact that $df\neq0$ on $E$ (in a mean value theorem?)?
Edit: I now have what I think is the solution.
Let $(a,b)\in E$. Then either $\frac{\partial f}{\partial x}(a,b)\neq 0$ or $\frac{\partial f}{\partial y}(a,b)\neq 0$. Assume without loss of generality that $\frac{\partial f}{\partial x}(a,b)\neq 0$. Then by the implicit function theorem, there exist open sets $U_{(a,b)}$ and $W_{(a,b)}$ such that $(a,b)\in U_{(a,b)}$, $b\in W_{(a,b)}$ and for every $y\in W_{(a,b)}$, there exists unique $x$ such that $(x,y)\in U_{(a,b)}$ and $f(x,y)=0$. We write $g_{(a,b)}(y)=x$, that is $f(g_{(a,b)}(y),y)=0$. By compactness, there exists $n\in \mathbb{N}$ such that we have $E\subseteq \cup_{i=1}^{n} U_{(a_{i},b_{i})}$.
Let $U_{i}=U_{(a_{i},b_{i})}, W_{i}=W_{(a_{i},b_{i})}$ and $g_{(a_{i},b_{i})}=g_{i}$.
Then $E=\cup_{i=1}^{n} U_{i}\cap E$. Let $(x,y)\in U_{i}\cap E$. Then $f(x,y)=0$ and $(x,y)\in U_{i}$. Observing the proof of the implicit function theorem, we see that this implies that $y\in W_{i}$. Hence $(x,y)=(g_{i}(y),y)$. That is, $U_{i}\cap E\subseteq \{(g_{i}(y),y):y\in W_{i}\}.$ For each $i$, $\mathcal{G}_{i}=\{(g_{i}(y),y):y\in W_{i}\}$ is a Jordan region of Jordan volume $0$ (as it is the graph of the function $g_{i}$). Hence $E\subseteq \cup_{i=1}^{n} \mathcal{G}_{i}$ is also a Jordan region with volume $0$.