I know that divergence of $\frac{\vec x}{x^3}$ is $4 \pi\delta^3(\vec x)$ and also that trace of gradient of a vector is its divergence.
When I take the gradient of $\frac{\vec x}{x^3}$ I get: $$\nabla\frac{\vec x}{x^3} = \frac{\mathbf I}{x^3} -3 \frac {\vec x \vec x}{x^5}$$
Which is traceless. So should I add $\frac {4 \pi}{3}\delta^3(\vec x) \mathbf{I}$ to the gradient?
Update: $4 \pi \delta^3(\vec x) \hat x \hat x$ also gives the correct trace. So it's more complicated now.