Vladimir's answer is off by factor of 2. The laplacian is $\nabla^2(\frac{1}{r^2}) = \frac{4}{r^4}$ A potential that falls of as $\frac{1}{r^2}$ is a dipole (In general, if it falls off as $r^{-n}$ its an ($2^{n-1}$)-pole, e.g $\frac{1}{r^3}$ bheaviour is quadrupole, etc).
Is this a dirac delta? To find out, check : $$\int_{\mbox{All space}}\nabla^2(\frac{1}{r^2})d^3r $$ $$=16\pi\int_0^\infty \frac{1}{r^2}dr\neq 1$$ Yup, integral diverges, so it is NOT a delta function.
I think your confusion is regarding the nature of a delta function. If something blows up at the origin it does not mean it is necessarily a delta function.
