is the inverse of a absolutely continuous function with almost everywhere positive derivation absolutely continuous?

It is always useful to have more than one answer to a problem. One learns quite a bit more.

Even better, however, is to have two contradictory answers to the same problem. That is both more entertaining and more educational. (The contradiction is just about what one wants to prove, not about errors.)

A friend of mine often tells an anecdote about a similar situation. A well-known mathematician had posed an open problem at the end of one of his papers. My friend sent him a solution. He wrote back quite pleased and informed my friend that someone else had also submitted a solution using a completely different method. He proposed that both should be published back-to-back in the Revue Roumaine to which, as an editor, he would submit them. And they were published. The methods were indeed completely different. My friend had solved the problem positively and the adjacent paper "proved" the opposite.

Problem. Suppose that $f:[0,1]\to\mathbb{R}$ is absolutely continuous and that $f'(x)>0$ for a.e. $x\in [0,1]$. Prove that $f$ has an absolutely continuous inverse.

Proof. Clearly $f$ is continuous and strictly increasing on $[0,1]$ and so it has a continuous and strictly increasing inverse $f^{-1}$. Let $P$ be the set of points at which $f$ has a finite, positive derivative. Let $Z$ be the remaining points which we know is a set of measure zero. Since $f$ is AC it satisfies Lusin's condition (N) so $f(Z)$ is also a set of measure zero. Let $E_1= f(P)$ and $E_2=f(Z)$. These sets exhaust $[f(0),f(1)]$. We know that $f^{-1}$ has a finite positive derivative at each point of $E_1$ and that $E_2$ has measure zero. The function $f^{-1}$ maps any measure zero subset of $E_1$ to a set of measure zero (because it has a finite derivative there). Also $f^{-1}$ maps every subset of $E_2$ to a set of measure zero (i.e., a subset of $Z$). Thus $f^{-1}$ satisifies Lusin's condition (N) on $[f(0),f(1)]$. It follows that $f^{-1}$ is absolutely continuous.

[Note added: The OP remembered a proof of this statement:

"Suppose $f'(x)$ exists at each point $x\in E$ and that $|f'(x)|\leq M$ there. Then $m(f(E))\leq Mm(E)$. Hence $f(E)$ is measure zero if $E$ is measure zero.

But this can be pushed. If $E$ has measure zero and $f'(x)$ is finite at every point of $E$ ($|f'(x)|$ not necessarily bounded) then simply write $E_n=\{x\in E: |f'(x)|\leq n\}$ and use the fact that $$m(f(E)) \leq \sum_{n=1}^\infty m(f(E_n))\leq \sum_{n=1}^\infty nm( E_n)=0.$$

In fact, with bit more work, one can prove that $$m(f(D)) \leq \int_D |f'(x)|\,dx$$ for any measurable set $D$ assuming that $f$ has a finite derivative at every point of $D$.

Let $c: [0,1] \to [0,1]$ be the usual Cantor-Lebesgue function and let $g: [0,1] \to [0,2]$ be given by $g(x) = c(x) + x$. Then $g$ is continuous, increasing, and $g'(x) = 1$ almost everywhere. Since $$g(1) - g(0) > \int_0^1 g'(x) \, dx$$ it follows that $g$ is not absolutely continuous.

Let $f = g^{-1}(t)$ so that $f : [0,2] \to [0,1]$ is continuous and increasing. If $x,y \in [0,2]$ and $x < y$ then $$\left| \frac{y-x}{f(y) - f(x)} \right| = \frac{y-x}{f(y) - f(x)} = \frac{g(f(y)) - g(f(x))}{f(y) - f(x)} = \frac{c(f(y)) - c(f(x))}{f(y) - f(x)} + \frac{f(y) - f(x)}{f(y) - f(x)}.$$ Since $c$ is nondecreasing, the last expression is greater than or equal to $1$. Thus $$|f(x) - f(y)| \le |x-y|.$$

It follows that $f$ is Lipschitz, hence AC, but $g = f^{-1}$ is not AC.