 Solution

For smooth compactly supported $u$, multiply by $1-\phi (nx)$. The removed term has $L^2$ norm $O(n^{-3/2})$, while its gradient terms are $O(n^{-1/2})$ in dimension three. Thus it tends to zero in $H^1$. Density then proves $H_0^1({\mathbb{R}}^3\setminus \{0\})=H^1({\mathbb{R}}^3)$.