Is $h(x)=g(x) \circ f^{-1}(x)$ twice differentiable when $g$ and $f$ is?

Question

I have some function $f$ that is continuous, strictly increasing and bijective (and therefore invertible) and twice continuously differentiable. Secondly, the function $g$ is also twice continuously differentiable. Consider then the function $h$:

$$h(x) = (g \circ f^{-1})(x)$$

How do I know if $h(x)$ is at least twice continuously differentiable?

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I gather from this answer that the smoothness of composite functions is equal to the smoothness of the least smooth function. Hence, $h(x)$ is at least twice continuously differentiable $\iff$ $f^{-1}(x)$ is twice continuously differentiable. Is this correct? If so, any suggestions on how to show that $f^{-1}$ is $C^2$ when $f(x)$ is $C^2$? Does the inverse function theorem help?

Answer 1

First claim: $h(x)$ is twice differentiable. Counterexample: $g(x) = x, \ f(x) = x^3$.

Second claim: $h(x)$ is twice continuously differentiable $\iff f^{-1}(x)$ is twice continuously diffentiable. Counterexample: $g(x) = f(x) = x^3$.

As a conclusion, the smoothness of the composition is at least the smoothness of the less smooth function, but can be greater.