Answer
$div E=0$ for the electric field $E(x) =\dfrac{\epsilon Q x}{|x^3|}$
Work Step by Step
$E(x) =\dfrac{\epsilon Q x}{|x^3|}$
This implies that $E(x)=\dfrac{\epsilon Q (xi+yj+zk) }{(x^2+y^2+z^2)^{3/2}}= \epsilon Q F$
Here, we have $f(x,y,z)=\dfrac{1}{(x^2+y^2+z^2)^{3/2}}$
Also, $\nabla f=f_xi+f_y j+f_z k$
Thus, we have
$\nabla f=-\dfrac{3x}{(x^2+y^2+z^2)^{5/2}}i-\dfrac{3y}{(x^2+y^2+z^2)^{5/2}} j-\dfrac{3z}{(x^2+y^2+z^2)^{5/2}}k$
or, $=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk)$
Therefore,
$div F=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk) \cdot (xi+yj+zk)+\dfrac{3}{(x^2+y^2+z^2)^{3/2}}=\dfrac{3}{(x^2+y^2+z^2)^{3/2}}-\dfrac{3}{(x^2+y^2+z^2)^{3/2}}=0$
This implies that $div E=0$ for the electric field $E(x) =\dfrac{\epsilon Q x}{|x^3|}$
