$$\bar{F} = \frac{\bar{r}}{r^3} = \frac{1}{r^3}(x\bar{i} + y\bar{j} + z\bar{k}) $$ $$\because (r = xi + yj + zk)$$ $$\bar{F} = r^{-3}x\bar{i} + r^{-3}y\bar{j} + r^{-3}z\bar{k}$$

Now by definition

Curl $\bar{F} = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ r^{-3} x & r^{-3} y & r^{-3} z \end{vmatrix} \\ \therefore \bar{F} = i\bigg[\frac{\partial}{\partial y} r^{-3}z - \frac{\partial}{\partial z}r^{-3}y\bigg] - j\bigg[\frac{\partial}{\partial x}r^{-3}z - \frac{\partial}{\partial z}r^{-3}x\bigg] + k\bigg[\frac{\partial}{\partial x}r^{-3} y - \frac{\partial}{\partial y}r^{-3}x\bigg] \\ \therefore \bar{F} = i\bigg[-3r^{-4}z\frac{\partial r}{\partial y} - 3r^{-4}y \frac{\partial r}{\partial z}\bigg] - j\bigg[-3r^{-4}z\frac{\partial r}{\partial x} + 3r^{-4}x\frac{\partial r}{\partial z}\bigg] + k\bigg[-3r^{-4}y\frac{\partial r}{\partial x} + 3r^{-4}x\frac{\partial r}{\partial y}\bigg]$

Now $r^2 = x^2 + y^2 + z^2$

Diff. w.r.t. x we get $2r\frac{dr}{dx} = 2x$

$\therefore \frac{dr}{dx} = \frac{x}{r}$

Similarly, $\frac{dr}{dy} = \frac{y}{r}, \frac{dr}{dz} = \frac{z}{r}$

Curl $\bar{F} = i\bigg[-3r^{-4}z\frac{y}{r} + 3r^{-4}y\frac{z}{r}\bigg] - j\bigg[-3r^{-4}z\frac{x}{r} + 3r^{-4}x\frac{z}{R}\bigg] + k\bigg[-3r^{-4}y\frac{x}{r} + 3r^{-4} x\frac{y}{r}\bigg] = 0$

$\therefore \bar{F}$ is irrotational

div $\bar{F} = \lfloor \bar{V} \cdot (r^{-3}xi + r^{-3}yj + r^{-3}zk)\rfloor \\ = \frac{\partial}{\partial x}(r^{-3}x) + \frac{\partial}{\partial y}(r^{-3}y) + \frac{\partial}{\partial z}(r^{-3}z) \\ = -3r^{-4}x\frac{\partial r}{\partial x} + r^{-3} - 3r^{-4}y\frac{\partial r}{\partial y} + r^{-3} - 3r^{-4}z\frac{\partial r}{\partial z} + r^{-3} \\ = 3r^{-3} - 3r^{-4} \frac{x^2}{r} - 3r^{-4}\frac{y^2}{r} - 3r^{-4}\frac{z^2}{r} \\ 3r^{-3} - 3r^{-5} \lfloor x^2 + y^2 + z^2 \rfloor \\ = 3r^{-4} - 3r^{-5}r^2 \\ = 3r^{-3} - 3r^{-3} \\ = 0$

$\therefore \bar{F}$ is solenoidal and irrotational.