This is an exercise of Introduction to the Theory of Distributions by Friedlander and Joshi.
Let $P$ be a linear differential operator with smooth coefficients defined on $\mathbb{R}^n$. Suppose that $P\delta=0$. Show that $P$ can be put into the form $$P=\sum_{j=1}^n P_j x_j,\quad \text{i.e.} Pu=\sum_{j=1}^n P_j(x_ju),$$ where $P_j$ are differential operators.
Here are my considerations. We use $P^T$ to denote the real transpose or adjoint of $P$. Then since $P\delta=0$ we see that $P^T\phi(0)=0$ for all test function $\phi$. Now we write by the fundamental theorem of calculus that $$P^T\phi(x) = P^T\phi(x)-P^T\phi(0)=\int_0^1 \frac{d}{dt}\left(P^T\phi(tx)\right)\mathrm{d}t=\int_0^1\sum_{j=1}^nx_j\frac{\partial P^T\phi}{\partial x^j}(tx)\mathrm{d}t.$$ We use the notation $$\mu_j \phi(x) = \int_0^1 \frac{\partial P^T\phi}{\partial x^j}(tx)\mathrm{d}t,$$ so now $$\langle Pu,\varphi \rangle = \langle u,P^T\phi \rangle=\langle u,\sum_{j=1}^nx_j\mu_j\phi\rangle=\langle\sum_{j=1}^n\mu_j^T(x_ju),\phi\rangle.$$
It remains to show that $\mu_j$are differential operators, but I am stuck at this place. One way to do so it to write the expression out with dilations and differentiations, which is complicated since it's not easy to find out how differentiations interact with dilations. Another way to prove the locality of $\mu_j$ i.e. $$\mathrm{supp} \mu_j\phi\subset \mathrm{supp}\phi,$$ which is also not very clear.
Or maybe, $\mu_j$ are not differential operators, but there do exist differential operators $P_j$ such that $$\sum \mu^T_jx_j=\sum P_jx_j.$$ But this possibility seems not operable at all.
Thanks for reading so far!