-Show $\nabla^2\left(\dfrac 1 {\mathbf{r}}\right)=0$
-Show $\operatorname{div}(f\mathbf{F})=f\operatorname{div}\mathbf{F}+\nabla f)\cdot \mathbf{F}$ And then show $\operatorname{div}\left(\dfrac {\mathbf{r} }{r^3}\right)=0$
I've tried but I actually don't know how to prove these.
First proof. $$\nabla^2 \frac{1}{r}= \nabla \cdot \nabla \frac{1}{r}$$ $$=\nabla \cdot \langle \frac{\partial}{\partial x} \frac{1}{r},\frac{\partial}{\partial y} \frac{1}{r}, \frac{\partial}{\partial z} \frac{1}{r} \rangle$$
$$=\nabla \cdot \langle \frac{\partial}{\partial x} \frac{1}{\sqrt{x^2+y^2+z^2}},\frac{\partial}{\partial y} \frac{1}{\sqrt{x^2+y^2+z^2}}, \frac{\partial}{\partial z} \frac{1}{\sqrt{x^2+y^2+z^2}} \rangle$$
$$=\nabla \cdot \langle \frac{-x}{\sqrt{(x^2+y^2+z^2)^3}},\frac{-y}{\sqrt{(x^2+y^2+z^2)^3}}, \frac{-z}{\sqrt{(x^2+y^2+z^2)^3}} \rangle$$
$$=\frac{\partial}{\partial x}\frac{-x}{\sqrt{(x^2+y^2+z^2)^3}}+\frac{\partial}{\partial y}\frac{-y}{\sqrt{(x^2+y^2+z^2)^3}}+\frac{\partial}{\partial z}\frac{-z}{\sqrt{(x^2+y^2+z^2)^3}} $$
$$=-1\left( (x^2+y^2+z^2)^{\frac{-3}{2}}-3x^2(x^2+y^2+z^2)^{\frac{-5}{2}}+(x^2+y^2+z^2)^{\frac{-3}{2}}-3y^2(x^2+y^2+z^2)^{\frac{-5}{2}}+(x^2+y^2+z^2)^{\frac{-3}{2}}-3z^2(x^2+y^2+z^2)^{\frac{-5}{2}} \right)$$
$$=-1\left( r^{-3}-3x^2r^{-5}+r^{-3}-3y^2r^{-5}+r^{-3}-3z^2r^{-5}\right)$$
$$=-1\left( 3r^{-3}-3r^{-5}(x^2+y^2+z^2)\right)$$
$$=-1\left(3r^{-3}-3r^{-5}r^2\right)=-1\left(3r^{-3}-3r^{-3} \right)=-1(0)=0$$
The third part is actualy part of this proof if you note $r^3=(x^2+y^2+z^2)^{\frac{3}{2}}$ then it is exactly the part after we took the gradient (first $\nabla$).