$\overrightarrow {r}= x\overrightarrow {i} + y \overrightarrow {j} + z\overrightarrow {k}$ $\therefore r^2 = x^2 + y^2 + z^2 \to (1)$ Partially differentiate w.r.t.x. $\dfrac {\partial}{\partial x}r^2 = \dfrac {\partial}{\partial x} (x^2 + y^2 + z^2)$ $\therefore 2r \dfrac {\partial r}{\partial x} = 2x$ $\therefore \dfrac {\partial r}{\partial x} = \dfrac {x}{r}$ Similarly, $\dfrac {\partial r}{\partial y} = \dfrac {y}{r}; \dfrac {\partial r}{\partial z} = \dfrac {z}{r} \ \ \ \to(2)$ $Now, \ \overline{F} = \dfrac {\overline r}{r^3}$ $ = \dfrac {x \overrightarrow {i} + y\overrightarrow {j}+ z \overrightarrow {k} }{r^3}$ $= \dfrac {x}{r^3} \overrightarrow{i} + \dfrac {y}{r^3} \overrightarrow {j}+ \dfrac {z}{r^3} \overrightarrow {k}$ Consider,  $\nabla \times \overline{F} \begin{vmatrix} \overrightarrow {i} & \overrightarrow {j} & \overrightarrow {k} \ \partial / \partial x & \partial / \partial y & \partial /\partial z \xr^{-3} & yr^{-3} &zr^{-3} \end{vmatrix}$ $=\overrightarrow {i} \left ( z\cdot -3r^{-4} \dfrac {\partial r} {\partial y} - y \cdot - 3r^{-4} \dfrac {\partial r}{\partial z}\right )- \overrightarrow {j} \left ( z\cdot -3r^{-4} \dfrac {\partial r}{\partial x}x\cdot - 3r^{-4} \dfrac {\partial r}{\partial z} \right ) + \overrightarrow {k} \left ( y \cdot - 3r^{-4} \dfrac {\partial r}{\partial x} - x\cdot - 3r^{-4}\dfrac {\partial r}{\partial y} \right )$ $= \overrightarrow{i}\left ( \dfrac {-3z}{r^4} \cdot \dfrac {y}{r} + \dfrac {3y}{r^4} \cdot \dfrac {z}{r} \right ) - \overrightarrow {j} \left ( \dfrac { -3z}{r^4}\cdot \dfrac {x}{r}+ \dfrac {3x}{r^4} \cdot \dfrac {z}{r} \right ) + \overrightarrow {k} \left ( \dfrac {-3y}{r^4} \cdot \dfrac {x}{r}+ \dfrac {3x}{r^4}\cdot \dfrac {y}{r} \right ) \ \ (from \ 2)$ $=0\overrightarrow {i} + 0 \overrightarrow {j} + 0\overrightarrow {k}$ $=0$ $\therefore \overline F \ is \ Irrotational$ Consider, $\nabla, \overline {F}= \left ( \overrightarrow {i}\dfrac {\partial }{\partial x} + \overrightarrow {j} \dfrac {\partial }{\partial y} + \overrightarrow {k} \dfrac {\partial }{\partial z}\right )\cdot \left ( \dfrac {x}{r^3} \overrightarrow {i}+ \dfrac {y}{r^3} \overrightarrow {j} + \dfrac {z}{r^3}\overrightarrow {k} \right )$  $= \dfrac {\partial }{\partial x}(xr^{-3}) + \dfrac {\partial }{\partial y} (yr^{-3})+\dfrac {\partial }{\partial z} (zr^{-3})$ $= \left ( x\cdot 3r^{-4} \dfrac {\partial r}{\partial x} + r^{-3}\cdot 1 \right )+ \left ( y\cdot -3r^{-4} \dfrac {\partial r}{\partial y}+ r^{-3}\cdot 1 \right ) + \left ( z\cdot -3r^{-4} \dfrac {\partial r}{\partial z}+r^{-3}\cdot 1 \right )$ $= -3r^{-4} x\cdot \dfrac {x}{r} + r^{-3} -3r^{-4}y\cdot \dfrac {y}{r}+r^{-3} -3r^{-4}z\cdot \dfrac {z}{r}+r^{-3} \ \ (from \ 2)$ $= -3r^{-5} (x^ 2+ y^2 + z^2) + 3r^{-3}$ $= 3r^{-5}(r^2) + 3r^{-3} \ \ (from \ 1)$ $=0$ $\therefore \ \overline F\ is \ Solenoidal$