How to show a function is not invertible locally? I have a silly question:
If a smooth function $f$  has a invertible derivative $Df(x_0)$ at a point $x_0$, then $f$ has a smooth inverse locally. This is what I learned in Inverse function Theorem. If we have a function $g$ such that $Dg(x_0)$ is not invertible, can we say that $g$ is not invertible locally?(Not necessarily smooth inverse)
Or in another words, what is the sufficient conditions to say a function is not invertible locally?

A sufficient condition for $f$ not to have a local inverse at $x_0$ is that $x_0$ is a local extremum.  For any $\delta$ small enough, $f(x_0+\delta), f(x_0 - \delta) \leq f(x_0)$, and hence we an order them as
$$f(x_0 \pm \delta) \leq f(x_0 \mp \delta) \leq f(x_0)$$
Hence, using the intermediate value theorem, we can show that $f$ is not invertible on $[x_0-\delta, x_0 + \delta]$.  Since $\delta$ can be taken to be arbitrarily small, this shows $f$ cannot be locally invertible.
Using the second derivative test, we can state this condition in terms of derivatives: if $f'(x_0) = 0$ and $f''(x_0) \neq 0$, then $f$ fails to be locally invertible at $x_0$.  However, if $f''(x_0) = 0$, the second derivative test fails, and $f$ may or may not be locally invertible (as the example $f(x) = x^3$ given in the comments shows).
