Let $f\in W^{1,p}(U)\,,U \subset\mathbb{R}^n$ open. If $f$ satisfies:
$$| D^\beta f(x) - D^\beta f(y)|\leq C |x-y|^\gamma $$
a. e. for some $0\leq \gamma \leq 1,\, x,y \in U$ and $|\beta|\leq 1$. Then there exists a subset $V$ of null measure such that $f$ is a differentiable function for $x\in U\setminus V$ and
$$ \partial_{x_i} f(x) = D_i f(x) \,\, a.e. x\in U$$
Here $D$ is the weak derivative and $\partial$ is the strong derivative.
I tried to start with the definition of a weak derivative, but I didn't go too far. Another attempt was to use the fundamental theorem of calculus at the beginning, but I arrive at an expression with the gradient of $ f $ and do not advance from there.
You can find the answered here. The concept of approximate differentability comes in handy here, since the approximate derivative is equivalent to the weak derivative for all $f\in W^{1,p}$.
Let $E \subset \mathbb{R}^{n}$ be a measurable set with finite measure. A function $f: E \rightarrow \mathbb{R}^{k}$ is approximately differentiable almost everywhere if for every $\varepsilon>0$ there is a compact set $F \subset E$ such that $\lambda(E \backslash F)<\varepsilon$ and $\left.f\right|_{F}$ is $C^{1}$ (i.e. there exists an extension $g$ of $\left.f\right|_{F}$ to $\mathbb{R}^{n}$ which is $C^{1})$. Clearly the Lebesgue measure of $E \backslash F$ is zero.
In the latter theorem it follows also that the classical differential of $\left.f\right|_{F}$ coincides with the approximate differential of $f$ at almost every $x_{0} \in F$.
For more details see Section 6 of L.C. Evans, R.F. Gariepy, "Measure theory and fine properties of functions" Studies in Advanced Mathematics.