My first question is: does there exist a smooth function $f$ such that $f \neq 0$ on $\mathbb{R}^n \setminus \{0\}$, $f(0) = 0$, and $1/f$, viewed as a distribution on $\mathbb{R}^n \setminus \{0\}$, can be extended to a distribution on $\mathbb{R}^n$, but nevertheless there does not exist any distribution $u \in \mathcal{D}'(\mathbb{R}^n)$ such that $fu = 1$?
At first I did not include the condition that "$1/f$ can be extended", but without it the problem is trivial: for instance, take $f(x) = e^{-1/x^2}$.
Łojasiewicz proved that if $f$ is a nonzero analytic function, then $1/f$ makes sense as an element of $\mathcal{D}'(\mathbb{R}^n)$. Treves gives a detailed account of this proof in Chapter 15 of Analytic Partial Differential Equations.
Naturally, this leads me to ask another question: does there exist a smooth function $f$ such that $f(0) = 0$, $f$ is not analytic at $0$, but there exists $u \in \mathcal{D}'(\mathbb{R}^n)$ such that $fu = 1$? As Gro-Tsen pointed out in the comments, $x^2 + e^{-1/x^2}$ is an example, since we can take $u = \frac{1}{x^2}\left(1 + \frac{e^{-1/x^2}}{x^2}\right)^{-1}$.
In fact, in §15.2.2 Treves discusses necessary and sufficient conditions for a distribution $u$ on $\mathbb{R}^n \setminus \{0\}$ to extend to a distribution on $\mathbb{R}^n$, namely whether $u$ can be regarded as an element of some negative-order Sobolev space $H^{-m}(B_R(0)\setminus \{0\})$. In particular, distributions of the form $1/x^k$ on $\mathbb{R} \setminus \{0\}$ can be extended to a distribution $u$ on $\mathbb{R}$, but this does not mean that $x^k u = 1$. In fact, $1 - x^k u$ must be a distribution supported at $\{0\}$, i.e. of the form $P(D)\delta$ for some differential operator $P(D)$.
In Generalized Functions (Vol. I, Ch. 1, §3), Gelfand gives an analytic continuation of the distributions $|x|^\lambda$ on $\mathbb{R}$ (see also Grafakos, Classical Fourier Analysis, §2.4.3 for a similar construction). When $\lambda$ is not a negative odd integer, the continuation is well-defined, while for negative odd integers $|x|^\lambda$ has poles. In particular, it seems that we cannot regard $|x|^{-1}$ as a distribution on $\mathbb{R}$.
Originally I wanted to ask whether there exists $u \in \mathcal{D}'(\mathbb{R})$ such that $|x| u = 1$. However, since $|x|$ is not a smooth function, the product with a distribution is not defined, which is why I turned to the first question. Nevertheless, I am still curious whether there is a reasonable way—without relying on analytic continuation—to phrase the statement “there is no distribution on $\mathbb{R}$ representing $|x|^{-1}$”. (Given Treves’ result that $|x|^{-1}$, viewed as a distribution on $\mathbb{R}\setminus \{0\}$, can always be extended to $\mathbb{R}$, it seems difficult to formulate such a statement clearly.)
I think my initial question is quite natural. As mentioned above, when $f$ is a nonzero analytic function or a polynomial, $1/f$ can always be realized as a distribution. I would like to know how deep this result is, how far it can be generalized, and whether there is a reasonably general theory for dealing with such smooth but non-analytic singularities.
