I was reading Rudin Functional Analysis
I came across the fact that even though the derivative of distribution exist in the classical sense, both weak derivative and classical derivative need not match.
Auther also provide counterexample for that 
I have the following doubt.
1) I do not find where in proof Auther proved f is absolute continuous
2) Also, I do not understand how to use Fubini theorem to equate both.
Please Help me.
Any Help will be appreciated
He doesn't prove or claim to have proved that $f$ is absolutely continuous. He shows that $f$ is AC if and only if the two derivatives are equal. (Indeed, the fact that $f$ may not be AC is the whole point: It's precisely when $f$ is not AC that the construction gives the counterexample he claims to be constructing.)
Maybe you're asking about the proof that $f$ is AC assuming the two derivatives are equal: We know from (1) that $D\Lambda_f=\Lambda_\mu$. So if $D\Lambda_f=\Lambda_{Df}$ then $\Lambda_{Df}=\Lambda_\mu$.. Hence $Df=\mu$, so $$f(x)-f(-\infty)=\mu((-\infty,x))=\int_{-\infty}^ Df(t)\,dt,$$which shows that $f$ is $AC$.
About the proof of (1) by Fubini: Suppose $\phi\in C^\infty_c(\Bbb R)$. Then $\newcommand\ip[2]{\langle #1,\rangle}$ $$\langle\phi,{D\Lambda_f}\rangle=-\int \phi' f =-\int_{-\infty}^\infty\phi'(t)\int_{-\infty}^t\,d\mu(x)dt =-\int_{-\infty}^\infty\int_x^\infty \phi'(t)\,dtd\mu(x)=\int\phi\,d\mu,$$which says precisely (1).