Let $F(x)=\int_0^xf(y)dy$ for all $x\in \mathbb{R}$, here the integral is a Lebesgue integral. If the function $f$ is continuous, then $F$ is continuously differentiable and has as derivative $f$.
Now if $f$ is only locally integrable, does $F$ has a weak derivative in the usual sense, that is, there exists a locally integrable function $G$ such that $\int_0^xF(y)\phi'(y)dy=-\int_0^xG(y)\phi(y)dy$ for all smooth compactly supported function $\phi$? in this case where can I find a proof of this result ?
Yes, and $f$ is the weak derivative.
Assume $f \in L^1(\mathbb{R})$ and let $F$ be its primitive function $$ F(x) = \int_{-\infty}^x f(y) dy = \int_{\mathbb{R}} f(y) \mathbf{1}_{y \le x} dy. $$ Choose any $\varphi \in C_c^\infty(\mathbb{R})$. By Fubini's theorem and the fundamental theorem of calculus, \begin{align*} \int_{\mathbb{R}} F(x) \varphi'(x) dx & = \int_{\mathbb{R}} \left( \int_{\mathbb{R}} f(y) \mathbf{1}_{y \le x} dy \right) \varphi'(x) dx \\ & = \int_{\mathbb{R}} \left( \int_{\mathbb{R}} \varphi'(x) \mathbf{1}_{x \ge y} dx \right) f(y) dy \\ & = - \int_{\mathbb{R}} \varphi(y) f(y) dy. \end{align*} If $f$ is merely locally integrable, one needs to extend it by zero outside the support of $\varphi$ (and possibly add a constant to $F$), but the main argument stays the same.
I see nothing wrong with approaches of @Pedro or @reuns, but this one seems the most straightforward to me.