# The zero of a continuously differentiable function is zero-measured

I'm tring to prove the following statement:

Suppose $$f:\mathbb{R}^2\rightarrow \mathbb{R}$$ is continuously differentiable, and for any $$(x_0,y_0)\in \mathbb{R}^2$$, we have $$\frac{\partial f}{\partial x}(x_0,y_0)+\frac{\partial f}{\partial y}(x_0,y_0)\neq 0$$ then show that: $$E=\{(x,y):f(x,y)=0\}$$ is a zero-measured set in $$\mathbb{R}^2$$.

It's easy to see that $$0$$ in the definition of $$E$$ can be replaced by any real number $$c$$. I have tried to integrate the two variable function $$\frac{\partial f}{\partial x}(x_0,y_0)+\frac{\partial f}{\partial y}(x_0,y_0)$$ and then applied the Fubini theorem but with no valuable findings. Can anybody give me some hints on proving this statement?

The desired conclusion can be obtained under the weaker assumption that $$f(x_0,y_0)=0 \; \Longrightarrow \; \frac{\partial f}{\partial x}(x_0,y_0)+\frac{\partial f}{\partial y}(x_0,y_0)\neq 0 \,.$$
For every real $$b$$, consider the function $$g_b:{\mathbb R} \to {\mathbb R}$$ defined by $$g_b(t)=f(t,t+b)$$. For every $$t$$, the chain rule and the hypothesis give $$g_b(t)=0 \; \Longrightarrow \; g_b'(t)=\frac{\partial f}{\partial x}(t,t+b)+\frac{\partial f}{\partial y}(t,t+b)\ne 0 \,,$$ so the zeros of $$g_b$$ are isolated, whence they form a countable set (at most). Thus the set $$E$$ of zeros of $$f$$ intersects every line of the form $$\{y=x+b\}$$ in a set of zero length, so Fubini's theorem implies that $$E$$ has zero area.
To formalize the last step, you may want to consider the (area preserving) rotation $$R$$ by $$45$$ degrees, observe that $$R(E)$$ intersects every vertical line in a set of zero length, and deduce that $$R(E)$$ (and hence also $$E$$) has zero area.