As per the title, I would like to prove that
$$\nabla\left\{\frac{1}{\|\cdot\|}\right\}=-\frac{1}{\|\cdot\|^3}x\qquad\text{in }\mathcal{D}'(\mathbb{R}^3).\qquad(\star)$$
Now, I am of the opinion that one would prove this in the following manner:
For all test functions $\phi\in\mathcal{D}(\mathbb{R}^3)$, we have
$$ \begin{aligned} \left\langle\nabla\left\{\frac{1}{\|\cdot\|}\right\},\phi\right\rangle&=-\left\langle\frac{1}{\|\cdot\|},\nabla\phi\right\rangle \\ &=-\lim_{\epsilon\downarrow 0}\int_{\|x\|>\epsilon}\frac{\nabla\phi(x)}{\|x\|}\,dx \\ &=\cdots \end{aligned} $$ and then evaluating the limits.
However, I was having a debate with a friend of mine, and he claimed that you could prove $(\star)$ in $C^\infty(\mathbb{R}^3\setminus\{0\})$ and then extend to distributions á la
$$\nabla\frac{1}{\|x\|}=-\frac{x}{\|x\|^3}.$$
Then you say that this holds in $\mathcal{D}'(\mathbb{R}^3)$, i.e.
$$\left\langle\nabla\left\{\frac{1}{\|\cdot\|}\right\},\phi\right\rangle=-\left\langle\frac{x}{\|\cdot\|^3},\phi\right\rangle$$
assuming one can show that $\nabla\left\{\frac{1}{\|\cdot\|}\right\}$ is a distribution. Is this way valid?
What your friend seems to want to say is this: consider $u \in \mathscr D ' (\Bbb R^3)$ such that $u = 0$ on $\Bbb R^3 \setminus \{0\}$; is it true then that $u=0$? Your concrete problem is just the particular case $u = \partial _i \frac 1 {\|x\|} + \frac {x_i} {\|x\|^3}$.
Well, this is not true: to see this just consider the case $u = \delta$ (Dirac's distribution). The deep reason why this extension attempt fails is because there is no natural way to uniquely extend distributions from an open set to an open superset. In general, the only non-ambiguous (functorial) set-theoretic operation that you can perform on a distribution is to restrict it to an open subset.