Answer
$div E=0$ for the electric field $E(x) =\dfrac{\epsilon Q x}{|x^3|}$
Work Step by Step
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, $S$ is a closed surface and $E$ is the region inside that surface.
$div F=\dfrac{\partial x}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial z}{\partial z}=1+1+1=3$
Here, $f(x,y,z)=\dfrac{1}{(x^2+y^2+z^2)^{3/2}}$
and $\nabla f=f_xi+f_y j+f_z k$
$\nabla f=-\dfrac{3x}{(x^2+y^2+z^2)^{5/2}}i-\dfrac{3y}{(x^2+y^2+z^2)^{5/2}} j-\dfrac{3z}{(x^2+y^2+z^2)^{5/2}}k=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk)$
Here, $E(x) =\dfrac{\epsilon Q x}{|x^3|}=\dfrac{\epsilon Q (xi+yj+zk) }{(x^2+y^2+z^2)^{3/2}}= \epsilon Q F$
Now, $div F=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk) \cdot (xi+yj+zk)+\dfrac{3}{(x^2+y^2+z^2)^{3/2}}=0$
Hence, it has been verified that $div E=0$ for the electric field $E(x) =\dfrac{\epsilon Q x}{|x^3|}$
