= Solution
By <locality of weak differentiability>, it is enough to work near the exceptional point; translate it to zero. Let $g_i$ denote the classical derivative off zero, assigning any value at zero. For $\phi\in C_c^\infty(\Omega)$ choose a smooth cutoff $\chi_\varepsilon$ which is zero on $B_\varepsilon$, one outside $B_{2\varepsilon}$, satisfies $|D\chi_\varepsilon|\leq C/\varepsilon$, and has its transition region contained in $\Omega$. The product $\phi\chi_\varepsilon$ avoids the singular point, so classical integration by parts gives
$$
\int u\chi_\varepsilon D_i\phi+\int u\phi D_i\chi_\varepsilon
=-\int g_i\phi\chi_\varepsilon.
$$
The middle term has absolute value at most
$$
\|u\|_\infty\|\phi\|_\infty\frac C\varepsilon |B_{2\varepsilon}|\leq C'\varepsilon^{n-1}\longrightarrow0
$$
when $n\geq2$. The other terms tend to $\int uD_i\phi$ and $-\int g_i\phi$, since $u$ is <locally integrable> and $g_i\in L^1_{\mathrm{loc}}$. Therefore $\boxed{D_i u=g_i\text{ weakly on all of }\Omega}$. Its value at the single point is irrelevant. This is <bounded isolated-point removability for weak first derivatives>.
For $n=1$, take $u(x)=0$ for $x<0$ and $u(x)=1$ for $x>0$ on $(-1,1)$, with any value at zero. It is bounded and $C^1$ off zero, with classical derivative zero there. But
$$
\int_{-1}^1u\phi'=-\phi(0),
$$
so its <distributional derivative> is a <Dirac delta>. If a <locally integrable> <weak derivative> existed, it would be zero off zero and hence zero <almost everywhere>, contradicting this identity. The step function is the required counterexample; the cutoff error in dimension one need not vanish.
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