Can a diffeomorphism have critical points?

I'm trying to understand the step highlighted in this demonstration:

from Zorich, Mathematical Analysis I, sec. 8.6, pag 510. What I know is that if $$f'(x_0)$$ is invertible ($$f:G\subset\mathbb{R}^m\to\mathbb{R}^m$$, $$f\in\mathcal{C}^p(G)$$), then f is locally a diffeomorphism (of smoothness $$p$$). In the step highlighted, instead, the opposite is stated, that is the other-sense implication ($$\Leftarrow$$). Is it always true? How would you prove it?

The map $$f$$ is not only differentiable. It is assumed that $$f$$ is a diffeomorphism. This means that $$f$$ is bijective, and that the inverse map $$g:=f^{-1}$$ is differentiable as well. Since $$g\bigl(f(x)\bigr)\equiv x$$ the chain rule implies that $$g'\bigl(f(x)\bigr)\circ f'(x)={\rm id}\ ,$$ hence $$f'(x)$$ has to be regular.