What are the real functions, defined on an interval, that decay under repeated differentiation? Let $f^{(n)}$ be the $n$-th derivative, and for all $n$ require $f^{(n)}$ to be continuous and not the zero function. Classify $f$ such that $f^{(n)}(x) \to 0$ for all $x$ as $n \to\infty$.
It seems that $e^{-1/x^2}$ satisfies this property. I am looking for references to general classes of such functions.
It turns out that every such function is of the form $f(x) = \sum_{n=0}^\infty a_n/n! \cdot (x-x_0)^n$ with $a_n \to 0$ and $x_0$ a point in the (open) interval on which $f$ is defined. Note in particular that this implies that the function is analytic; thus, your function $e^{-1/x^2}$ does not satisfy your condition.
Note: I am not sure currently whether each function of the above form actually satisfies your condition. I still have to think about this.
Let us show that each function satisfying your condition is of the claimed form.
To this end, note that the Taylor series around $x_0$ of $f$ is given by $\sum_{n=0}^\infty f^{(n)}(x_0)/n! \cdot (x-x_0)^n$. The radius of convergence of this power series is given by $\varrho(x_0) = 1 / \limsup_{n\to\infty} \sqrt[n]{|f^{(n)}(x_0)/n!|} = \infty$, since by your assumption you have $|f^{(n)}(x_0)| \leq C$ for some $C = C(x_0) > 0$, and therefore $\sqrt[n]{|f^{(n)}(x_0)|/n!} \leq \sqrt[n]{C} / \sqrt[n]{n!} = 0$, since $\sqrt[n]{n!} \to \infty$ and $\sqrt[n]{C} \to 1$.
Now, a Theorem of Pringsheim (which is only correctly proven by Boas, see https://www.johndcook.com/blog/2016/10/25/when-does-a-function-equal-its-taylor-series/) implies in particular that if $\varrho(x_0) \geq c > 0$ for all $x_0$, then $f$ is in fact analytic. This proves that $f$ is analytic.
Now, let $x_0 \in I$ be arbitrary. Since $f$ is analytic (as we just saw), and since $\varrho(x_0) = \infty$, it follows that $f(x) = \sum_{n=0}^\infty f^{(n)}(x_0) / n! \cdot (x-x_0)^n$ for all $x \in I$, so that the condition that I claimed above holds for $f$, if we define $a_n := f^{(n)}(x_0)$.