Is a C^1 diffeomorphism with f of class C^k also a C^k diffeomorphism?

[Avogadro Number]

I am studying Choquet-Bruhat's Introduction to General Relativity, Black Holes and Cosmology, and I don't follow the hint in Exercise I.2.1:

Exercise I.2.1 Let U,V be open subsets of R^d. Prove that a C^1 diffeomorphism f:U-->V with f of class C^k is a C^k diffeomorphism.

Hint: $\partial (f f^{-1})/ \partial x^i \equiv 0$.

Isn't f f^{-1} the identity map, and then how is the claimed partial identically zero? And how is this useful? Are we then meant to use the Chain Rule?
I would be grateful for any help! Thanks!

[Moderator's note: Moved from SR/GR.]

 

[Avogadro Number]

Actually, I now realize that the Chain Rule does help, and probably the "0" in the hint is a typo, and is probably meant to be an identity, with the partial derivative replaced by derivative of as a map from R^d to R^d.
Then f o f^{-1}=id , the Chain Rule , and the Inverse Function Theorem can be used to show that f^{-1} is C^k too.

 

[WWGD]

All I can think of is that if f is a diffeomorphism, then $f(f^{-1}(x))=x$. But this is for a vector $x=(x_1,...,x_d)$

 

[Avogadro Number]

As $f\circ f^{-1}=\textrm{id}$, we obtain $(Df)(f^{-1}(x))\cdot (Df^{-1})(x)=\textrm{id}$, and so we see that $(Df)(f^{-1}(x))$ is an invertible matrix, and also $(Df^{-1})(x)=((Df)(f^{-1}(x)))^{-1}$, and by Cramer's Rule,
we see that the $(i,j)$th entry of $Df^{-1}(x)$ is given by a polynomial combination of the partials of $f$, divided by the nonzero Jacobian determinant of $Df$. As both of the latter are $C^{k-1}$, it follows that the partials of $Df^{-1}$ are also $C^{k-1}$, and so $f^{-1}$ is $C^k$.