Let $u(x) \in H^1(\mathbb R^n;\mathbb C)$ be a complex-valued Sobolev function. One can check that $v(x) = |u(x)| \in H^1(\mathbb R^n;\mathbb C)$ with $$\nabla |u(x)| = \frac{\Re(u \nabla u)}{|u(x)|}$$ (defined as zero when $u(x) = 0$) by considering the sequence $$v_{\varepsilon} = \sqrt{|u(x)|^2 + \varepsilon}$$ which converges to $|u(x)|$ in $H^1(\Omega)$ for any open and bounded $\Omega \subset \mathbb R^n$.
Now, let $$w(x) = \frac{u(x)}{|u(x)|}$$
If I were to apply the chain-rule without justification, I would find $$\nabla w(x) = \frac{\nabla u(x) |u(x)|- u(x) \nabla|u(x)| }{|u(x)|^2}$$ But to apply the chain-rule, I need some kind of Lipschitz hypothesis on the nonlinear map that I'm applying to $u(x)$, which we don't have here I believe.
I've tried to justify the existence of this weak derivative by considering $$w_{\varepsilon} = \frac{u(x)}{\sqrt{|u(x)|^2 + \varepsilon}}$$ Maybe my computations were wrong, but I couldn't prove that $\nabla w_{\varepsilon}$ has a limit in an appropriate $L^p$ space.
No, generally it doesn't have a weak derivative in $L^1_{\mathrm{loc}}$.
In one dimension this is clear, because if $w$ has a weak derivative in $L^1_{\mathrm{loc}}$ then it is continuous, which $w = \frac{u}{|u|}$ typically is not.
In higher dimensions you can get a counterexample by letting $u(x_1, \dots, x_n) = u_0(x_1)\psi(x_1, \dots, x_n)$ where $u_0$ is a counterexample in $\mathbb{R}^1$ and $\psi$ is an appropriate cutoff function.