Let $\Omega$ be the open square $(-1,1)^2\subset\mathbb{R}^2$, $u\in L^1(\Omega)$ such that for each $(x,y)\in\Omega$, $u(x,y)=1-\max\{|x|,|y|\}$.
Determine the weak gradient of $u$ and find $p$ such that $u\in W^{1,p}(\Omega)$.
Given $\phi\in\mathcal{D}(\Omega)$, I need to compute separately weak $\partial_xu$ and weak $\partial_yu$: $$\langle\partial_xu,\phi\rangle = -\langle u,\partial_x\phi\rangle = -\int_\Omega \left(1-\max\{|x|,|y|\}\right)\partial_x\phi\,dx\,dy.$$ I can split the domain $\Omega$ in $$\begin{aligned} A&=\{(x,y)\in\Omega:y>|x|\}, &B&=\{(x,y)\in\Omega:x>|y|\},\\ C&=\{(x,y)\in\Omega:y<-|x|\}, &D&=\{(x,y)\in\Omega:x<-|y|\}, \end{aligned}$$ but I'm having difficulties in integrating by parts.
Following the suggestion in the comments to simply use Fubini-Tonelli's Theorem (since $u\,\partial_x\phi\in L^1(\Omega)$):
$$ \int_\Omega u(x,y)\partial_x\phi(x,y)\,dx\,dy = \int_{-1}^1\underbrace{\int_{-1}^1u(x,y)\partial_x\phi(x,y)\,dx}_{\displaystyle I(y)}\,dy \\ \begin{aligned} I(y) &= \int_{-1}^1(1-\max\{|x|,|y|\})\partial_x\phi(x,y)\,dx = \dots \\ &= -\int_{-1}^{-|y|}\phi(x,y)\,dx + \int_{|y|}^1\phi(x,y)\,dx \\ &= \int_{-1}^1\left[\pmb{1}_{(|y|,1)}(x)-\pmb{1}_{(-1,|y|)}(x)\right]\phi(x,y)\,dx, \end{aligned} $$ where $\pmb{1}_S$ is the indicator function of the set $S$. So we simply have $$ \begin{aligned} \langle\partial_x u,\phi\rangle &= -\int_\Omega u(x,y)\partial_x\phi(x,y)\,dx\,dy \\ &= -\int_\Omega\left[-\pmb{1}_D+\pmb{1}_B\right]\phi\,dx\,dy = \int_\Omega\left[\pmb{1}_D-\pmb{1}_B\right]\phi\,dx\,dy \end{aligned} $$ and thus the weak partial $x$-derivative of $u$ would be $\pmb{1}_D-\pmb{1}_B$. Is that correct?