Let $f: [0,1] \rightarrow \mathbb{R}$ be a $C^1$ function. By the mean value theorem, we have, for every $x,y \in [0,1]$ the following inequality
$$|f(x) - f(y)| \leq \max_{0\leq z \leq 1}|f'(z)||x-y|\ .$$
Is it true that, if we have $x < y \in [0,1]$ such that equality holds, then $$f'(\xi) = \max_{x\leq z \leq y}|f'(z)|, \quad \forall \xi \in [x,y] \ ,$$
that is, $f'$ is constant on $[x,y] \subset [0,1]$?
On
Suppose you have equality for $0 \le x < y \le 1$. Multiplying by $-1$ if necessary we can assume that $f(y) > f(x)$. Write $M = \max_{x\le z \le y} |f'(z)|$. Then $$ |f(y) - f(x)| = f(y) - f(x) = \int_x^y f'(z) dz = \int_x^y (f'(z) -M)dz + \int_x^y M dz \\ = \int_x^y (f'(z) -M)dz + M|y-x|. $$ Due to the assumed equality we then know that $$ 0 = \int_x^y (f'(z) -M)dz, $$ but since $f' - M$ is continuous and nonpositive we deduce that $f'(z) = M$ for $z \in (x,y)$ and hence that $f$ is linear in this region.
You can use Taylor-integral formula : $f(y)=f(x)+\int_x^y f'(t)dt$
Now let's call $M=\max\limits_{x\le z\le y}|f'(z)|$ and suppose we have $\frac{f(y)-f(x)}{y-x}=M$ for some $x,y$.
Then we have $0=f(y)-f(x)-\int_x^y f'(t)dt=M(y-x)-\int_x^y f'(t)dt=\int_x^y \big(M-f'(t)\big)dt$
But $M-f'(t)\ge 0$ for any $t\in[x,y]$ by definition of $M$.
We conclude that the integral can be null only if $f'(t)=M$ almost everywhere on $[x,y]$ (i.e. $\forall t\in[x,y]\setminus A$ where $\mu(A)=0$)
But you stated that $f$ is $C^1$ so by continuity $f'$ is constant on whole $[x,y]$.