Why is the differential being onto equivalent to it not being zero?

[yifli]

I have difficulty understanding the following Theorem

If U is open in $ℝ^2$, $F: U \rightarrow ℝ$ is a differentiable function with Lipschitz derivative, and $X_c=\{x\in U|F(x)=c\}$, then $X_c$ is a smooth curve if $[\operatorname{D}F(\textbf{a})]$ is onto for $\textbf{a}\in X_c$; i.e., if $$\big[ \operatorname{D}F\bigl( \begin{smallmatrix}a \\ b\end{smallmatrix}\bigr)\big]≠0 \mbox{ for all } \textbf{a}=\bigl( \begin{smallmatrix}a \\ b \end{smallmatrix}\bigr)\in X_c$$

I don't understand why the differential of F at a being onto is equivalent to saying the differential is not zero. Can someone explain? Thanks

 

[HallsofIvy]

Think about what happens for simple functions like $f(x)= x^2$ where f'(x)= 0.

 

[yifli]

Think about what happens for simple functions like $f(x)= x^2$ where f'(x)= 0.

the differential of $f(x)=x^2$ is $2x$, so $f'(x)=0$ means x=0.
Now what?