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