Divergence of "inverse square law" vector field 
Let $\vec{F}$ be an inverse square law vector field given by
  $$
\vec{F} = \frac{\vec{r}}{\lVert \vec{r} \rVert^3} 
$$
  where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Prove $\nabla \cdot \vec{F} = 0$.

If I can show $\vec{F} \cdot \hat{n} = 0$, then
$$
\oint \vec{F} \cdot \hat{n}dS = 0
$$  
Then apply the Divergence Theorem,
$$
\oint \vec{F} \cdot \hat{n} \ dS = \iiint\nabla \cdot \vec{F} \ dV
$$
and so 
$$
\iiint\nabla \cdot \vec{F} \ dV = 0 \implies \nabla \cdot \vec{F} = 0
$$
The part I'm stuck on is proving $\vec{F} \cdot \hat{n} = 0$. Can someone point me in the right direction?

Recall that if $\varphi$ is a scalar field and $\vec{G}$ is a vector field, then: $$\nabla \cdot (\varphi \vec{G}) = \varphi \,\nabla \cdot \vec{G} + (\nabla \varphi) \cdot \vec{G}.$$Make $\varphi = 1/\|\vec{r}\|^3 $ and $\vec{G} = \vec{r}$. Writing $$\varphi(x_1,x_2,x_3) = (x_1^2+x_2^2+x_3^2)^{-3/2},$$we get $$\partial_i\varphi(x_1,x_2,x_3) = -3x_i(x_1^2+x_2^2+x_3^2)^{-5/2} \implies \nabla \varphi = -3\frac{\vec{r}}{\|\vec{r}\|^5}$$So: $$\nabla \cdot\vec{F} = \frac{1}{\|\vec{r}\|^3}3 - 3\frac{1}{\|\vec{r}\|^5}\vec{r}\cdot\vec{r} = \frac{3}{\|\vec{r}\|^3}-\frac{3}{\|\vec{r}\|^3} = 0,$$painlessly.
