I have $x \cdot \partial_x \delta(x-y)$, and want to simplify it. I believe it works like this: I use $x$ as a test-function and use integration by parts $$\int_{-\infty}^{\infty} x \cdot \partial_x \delta(x-y) dx = \delta(x-y) x - \int_{-\infty}^{\infty} \delta(x-y) dx $$
As x is anti-symmetric, the first part cancels, and we get
$$x \cdot \partial_x \delta(x-y)=-\delta(x-y)$$
Is that true?
For a test function $\phi \in \mathcal D(\mathbb R)\def\R{\mathbb R}$ and your $T = x \cdot \partial_x(\tau_{y}\delta)$, we have$\def\<#1>{\left<#1\right>}$ \begin{align*} \<\phi, T> &= \<\phi, x\tau_{y}\partial_x\delta>\\ &= \<x\phi, \tau_{y}\partial_x\delta)>\\ &= \<(x+y)\tau_{-y}\phi, \partial_x\delta>\\ &= -\<\tau_{-y}\phi + (x+y)\tau_{-y}\phi', \delta>\\ &= -\phi(y) - y\phi'(y)\\ &= \<\phi, -\tau_{y}\delta + y\tau_y\partial_x\delta> \end{align*} So, in the your notation, we have $$ x\cdot \partial_x\delta(x-y) = -\delta(x-y) + y\partial_x\delta(x-y) $$
Notation: $\tau_y$ denotes the shift by $y$, that is $\tau_y\phi(x) = \phi(x-y)$.
