Exercise 4.4.5: (required 4.3) Show that if in the inverse function theorem $f$ has $k$ continuous derivatives, then the inverse function $g$ also has $k$ continuous derivatives.
We know that $$g'(x) = \frac1{f'(g(x))},$$
by the inverse function theorem. Therefore, it is true that if $f$ has $k$ continuous derivatives, so does $g$. However, I am struggling with proving this. Section 4.3 is for Taylor series, and I don't know how this helps this question. I appreciate if you give some help.
It seems that you are talking about one-variable functions.
We need the following Lemma: If $f$, $g\in C^r$, and $g\circ f$ is defined, then $g\circ f\in C^r$.
Proof. This is true for $r=0$. Assume it for $r-1$. Then the chain rule gives $$(g\circ f)'=(g'\circ f)\cdot f'\in C^{r-1}\cdot C^{r-1}\subset C^{r-1}\ ,$$ where the $\in$ comes from the induction hypothesis and the $\subset$ from Leibniz' formula for the higher derivatives of a product.$\quad\square$
Your formula says that $$g'(y)={1\over f'\bigl(g(y)\bigr)}$$ in a full neighborhood $V$ of the point $y_0:=f(x_0)$. This means that $$g'=\iota\circ f'\circ g\ ,\tag{1}$$ where the map $\iota:\> t\mapsto {1\over t}$ is just taking reciprocals. The Lemma says that $g'$ is at least as often differentiable as any "factor" on the RHS of $(1)$. As $\iota\in C^\infty$ and $f'\in C^{r-1}$ you can set up a full induction proof leading to $g'\in C^{r-1}$.