It's quite easy to see that we can multiply distributions by any $\mathcal C^\infty $ functions. Moreover, if the distribution $T$ is of order $k$, then we can mupliply it by a $\mathcal C^k$ function.
The problem is, what to do with division by a function. For example, consider the equation $$|x|S =0,\quad S\in \mathcal D'(\Bbb R).$$ One can show that the support of $S$ is a subset of $\{0\}$. Clearly, we can solve this equation in the class of distributions of order $0$ (easy to see that the solution is $Const\cdot \delta_0$). Then again, for derivatives of $\delta_0$ the expression $|x|\delta_0^{(k)}$ is not even defined (more precisely, it's defined on a subspace of $\mathcal D(\Bbb R)$, but not the whole $\mathcal D(\Bbb R)$). Therefore, we can say that in a certain way we "solved" this incorrectly posed problem.
On the other hand, the equations $|x|S=1$, $|x|S=\delta_0$ don't have solutions (up to some mistake I could have done=)).
So, the question is, is there a criterion for saying that the equation $$|x|S=T$$ has/doesn't have solutions in the class of distributions? And, more generally, if we solve $f(x)S=T$ for $f\in \mathcal C^k\setminus\mathcal C^\infty$?