A continuously differentiable bijection implies its inverse is Lipschitz continuous

Question

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuously differentiable bijection. Does this imply that $f^{-1}$ is Lipschitz continuous? (of course, not globally, take for instance $f(x)=x^3$)

If not, what if $f\in C^2$ in addition?

I tried stupid things like... Fix an interval $[a,b]$ and pick $x,y\in [a,b]$. Then $$|x-y|= |f(f^{-1}(x)) - f(f^{-1}(y))|\leq \|f'\|_{\infty} |f^{-1}(x)-f^{-1}(y)|$$ where $\|f'\|_{\infty} := \sup_{x\in [a,b]} |f'(x)|$. But I don't see how to conclude from this.

Thanks!

Answer 1

Doesn't $x^3$ provided your counterexample?

$f^{-1}(y ) = y^{1/3}.$

So

$$ \frac{f^{-1}(y) - f^{-1}(0)}{y-0} = y^{-2/3}, $$ which is not less than a constant times $y.$ (I am just looking at $y>0$ but that's all that needed here.)