Hi, I was trying to solve this question. If we assume that the field is defined in space, the divergence becomes 0 which makes sense(even though the function is expanding) because the field is not defined at the origin. But if we assume that the field is defined in the $xy$ plane(which actually isn't what the question intended), the divergence becomes negative.($\dfrac{-1}{r^3}$)
I don't understand why this happens. The function seems to be expanding geometrically so the divergence should be positive.
This is how I computed: $\dfrac{-1}{r^3}$:
$v = \dfrac{1}{(x^2 + y^2)^{\dfrac{3}{2}}} \times (xi + yj)$
$=> \nabla . v = \dfrac{2r^3 - 3r^3}{r^6} = \dfrac{-1}{r^3}$
It is true that $\nabla \cdot \left ( \frac{\mathbf{r}}{r^3} \right ) < 0$ if $\mathbf{r} \neq \mathbf{0}$ in 2D. But it is a positive delta function at the origin. Consequently the integral of it over a disk centered at the origin is still positive, in accordance with the outward character of the field. This negativity is saying that the flux through a circle decays as the radius of the circle grows. This can be viewed as a consequence of the fact that in 3D the surface area of the sphere is proportional to $r^2$ while in 2D it is proportional to $r$ (so the flux behaves like $1/r$ which decays). This fact also shows up in the difference between the fundamental solutions to the Laplace equation in 2D and 3D.