Let's consider $f:\mathbb{R}\to\mathbb{R}$, with
$$f(x)=\begin{cases}e^{-1/x^2}&x\neq 0\\ 0&x=0. \end{cases}$$ The $n$-th derivative exists and is given by $f^n(x)=x^{-3k}P_n(x)f(x)$ if $x\neq 0$ and if $x=0$ then the derivative for all $n$ is $0$. $P_n(x)$ is a polynomial of $n$-th degree.
For example: $$ f'(x)=x^{-3}2f(x)\\ f''(x)=x^{-6}(4-6x^2)f(x)\\ \cdots $$ How do I show that there exists no $K\in\mathbb{R}$ such that $|f^n(x)|\leq K$ for all $n\in \mathbb{N}$?
EDIT:
My approach:
The $n$-th derivative takes the form of $f^{n}(x)=x^{-3n}e^{-1/x^2}P_n(x)$ where $P_n(x)$ is a polynomial of $n$-th degree. The function $h(x):=x^{-3n}e^{-\frac{1}{x^2}}$ has a maximum at point $\sqrt{\frac{2}{3n}}$ such that $h\left(\sqrt{\frac{2}{3n}}\right)=\left(\frac{3n}{e^1 2}\right)^{\frac{3n}{2}}$. This a strictly increasing function with respect to $n$ and hence unbounded. Further, if I plug in $\sqrt{\frac{2}{3n}}$ into $P_n(x)$ one can notice that if $n\to\infty$ the polymomial $P\left(\sqrt{\frac{2}{3n}}\right)$ goes to $a_0$, the summand of the polynomial which has no $x$-variable. So the $n$-th derivative $f^{n}(x)=x^{-3n}e^{-1/x^2}P_n(x)$ is unbounded.
No, there is no such uniform bounded, otherwise $f$ would be real analytic, see the $3$-rd alternative characterization here https://en.wikipedia.org/wiki/Analytic_function#Alternative_characterizations
Now, you know that $f$ has no Taylor series expansion near $0$ as, $f^{(n)}(0)=0$ for all $n\geq 1$, but $f\not \equiv 0$ near $0$.
Alternative Way: If possible let $||f^{(n)}||_\infty<M$ for some $M$ and for all $n\geq 1$. Let $0<t\leq 1$. Then, $\big|f^{(k-1)}(t)\big|\leq \int_0^t\big|f^{(k)}(s)\big|\ ds\leq Mt$ and $\big|f^{(k-2)}(t)\big|\leq \int_0^t\big|f^{(k-1)}(s)\big|\ ds\leq \frac{Mt^2}{2!}$, and inductively $\big|f(t)\big|\leq \frac{Mt^k}{k!}\leq \frac{M}{k!}\to 0$ as $k\to\infty$, which forces $f\equiv 0$ on $(0,1)$, contradiction.