Prove that if $f^{(k)}(f^{-1}(x))$ exists, and is nonzero, then $(f^{-1})^{(k)}(x)$ exists. I think I found an error and just want to confirm.
This is from Calculus by Michael Spivak, 3rd edition, Chapter 12 (Inverse Functions), problem 21.

12-21. Prove that if $f^{(k)}(f^{-1}(x))$ exists, and is nonzero, then $(f^{-1})^{(k)}(x)$ exists.

This appears to be false. For example, take $f(x) = x^3$.
In this case
$$f^{-1}(x) = x^{\frac{1}{3}}$$
$$f(0) = f^{-1}(0) = 0$$
$$f^{(3)}(0) = 6$$
If the statement is true, $(f^{-1})^{(3)}(0)$ would exist. However, $(f^{-1})'(0)$ doesn't exist (because $f'(0) = 0$) so higher order derivatives cannot.
I think the question should instead say "Prove that if $f^{(k)}(f^{-1}(x))$ exists, and $f'(f^{-1}(x))$ is nonzero, then $(f^{-1})^{(k)}(x)$ exists.
Does that sound about right to y'all?
Edit: As discussed with Paul Frost, the revised version of the statement

Prove that if $f^{(k)}(f^{-1}(x))$ exists, and $f'(f^{-1}(x))$ is nonzero, then $(f^{-1})^{(k)}(x)$ exists

is missing one other thing.
We need $f$ to be one-one on some interval containing $f^{-1}(x)$.
For cases $k\geq 2$, $f$ one-one is implied by $f'(f^{-1}(x))≠0$, and the existence of $″(f^{-1}(x))$ (which gives us the existence of $f'$ for all points in some interval containing $f^{-1}(x)$ and continuity of $f'$ at $f^{-1}(x)$). We can prove the (corrected) statement for $k≥2$. However, we don't know if $f$ is one-one for the case $k=1$. As problem 11-63 shows, $f'(f^{-1}(x))\neq 0$ by itself isn't enough to guarantee $f$ is increasing (or decreasing) over some interval containing $f^{-1}(x)$
But that's ok. The higher derivative cases are the ones we're interested in anyway, so we can take $k = 2$ as the base case and use induction from there.
The corrected corrected statement reads:
"Prove that if $f^{(k)}(f^{-1}(x))$ exists, and $f'(f^{-1}(x))$ is nonzero, then $(f^{-1})^{(k)}(x)$ exists for all $k\geq 2$, $k\in \mathbb{N}$
For the $k=1$ case, we have theorem 12-5.
 A: Your counterexample is correct; Spivak seems to be a bit unattentive. He also does not say anything about $f$ except that $f^{(k)}(f^{-1}(x_0))$ exists (and is $\ne 0$, which you have shown to be inadequate). The existence of $f^{(k)}(f^{-1}(x_0))$ means more precisely that (a) there is an open neigborhood $U$ of $f^{-1}(x_0)$ on which $f$ is $(k-1)$-times differentiable, and (b) $f^{(k)}(f^{-1}(x_0))$ exists.
In Theorems 12.4 and 12.5 Spivak states for an injective function $f$ which is differentiable at $f^{-1}(x_0)$:

*

*If $f'(f^{-1}(x_0)) = 0$, then $f^{-1}$ is not differentiable at $x_0$.


*If $f'(f^{-1}(x_0)) \ne 0$, then $f^{-1}$ is differentiable at $x_0$ and $(f^{-1})'(x_0) = \dfrac{1}{f'(f^{-1}(x_0))}$.
Thus the minimal requirement for the existence of $(f^{-1})'(x_0)$, and hence for the existence of $(f^{-1})^{(k)}(x_0)$ for some $k \ge 1$, is
$$f'(f^{-1}(x_0)) \ne 0 \tag{1}.$$
Now assume that $f^{(k)}(f^{-1}(x_0))$ exists (with any value of $f^{(k)}(f^{-1}(x_0))$) and $(1)$ is satisfied. We shall (a) construct an open neigborhood $V$ of $x_0$ such that $f^{-1}$ is $(k-1)$-times differentiable on $V$ (b) show that $(f^{-1})^{(k)}(x_0))$ exists.
For $k = 1$ nothing remains to do; 2. applies. We now consider $k > 1$.
Since $f''(f^{-1}(x_0))$ exists, $f'$ is continuous at $f^{-1}(x_0)$, thus $f'(y) \ne 0$ for all $y$ in some open neigborhood $U' \subset U$ of $f^{-1}(x_0)$. For $x \in V = f(U')$ we have $f^{-1}(x) \in U'$, thus by 2. $f^{-1}$ is differentiable at $x$ with derivative  $(f^{-1})'(x) =  \dfrac{1}{f'(f^{-1}(x))}$. Clearly $x_0 = f(f^{-1}(x_0)) \in V$. Since $f^{-1}$ is continuous (Theorem 12.3), $V = f(U') = (f^{-1})^{-1}(U')$ is open. Thus $V$ is an open neigborhood of $x_0$.
We shall now show by induction on $r$ that $(f^{-1})^{(r)}(x_0)$ exists on $V$ for all $r < k$. The base case $r=1$ has been treated above. Let us now consider $r=2$, the details of the general case are left to you.
Quotient rule and chain rule show that $(f^{-1})''(x)$ exists for all $x \in V$, where
$$(f^{-1})''(x) = -\dfrac{f''(f^{-1}(x))}{(f'(f^{-1}(x)))^3} .$$
Induction produces for all $x \in V$ a formula for $(f^{-1})^{(r)}(x)$ as a fraction whose denominator is a power of $f'(f^{-1}(x))$ and whose nominator has summands which are products of powers of derivatives $f^{(i)}(f^{-1}(x))$ with $i = 1,\ldots,r$.
This shows that  $(f^{-1})^{(k-1)}(x_0)$ exists on $V$. Clearly we can once more differentiate at $x_0$.
