Showing that this function is infinitely differentiable Show, that the function
$$ \mathcal E: \mathbb R \to \mathbb R:
 x \mapsto \begin{cases}
 \exp(-\frac{1}{x^2}), & \text{if x $\neq$ 0}, \\
 0, & \text{otherwise},
 \end{cases} $$
is infinitely differentiable and that $\frac{d^k\mathcal E}{dx}(0) = 0$ for all $k \in \mathbb N$.
We just introduced differentiation, so the solution should not contain very advanced techniques to solve this. We also got the tip it should be done by induction.
@fgp I found that it is
$$f^{(n)}(x) = P_n \left(\frac1x\right)e^{-\frac1{x^2}}$$
where $P_n$ is a polynomial with integer coefficients.
I could also do the initial step for the induction:
For $n=1:$
$$f'(x) = \frac2{x^2}e^{-\frac1{x^2}} = P_1 \left(\frac1t\right) e^{-\frac1{x^2}}$$
where $P_1(x)=2x^2.$
I am stuck at the induction step: $f^{(n+1)}(x) = $.....
 A: (This is more a hint than a full answer) Note that 
$$
\begin{align*}
\frac{d\mathcal{E}}{dx}(0) &= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} \frac{\mathcal{E}(x)-\mathcal{E}(0)}{x} \\
&= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} \frac{1}{x} \exp \left( -\frac{1}{x^2} \right) \\[2mm]
&= 0
\end{align*}$$
So, $\mathcal{E}'(0)=0$. As well, we have :
$$
\begin{align*}
\frac{d^{2}\mathcal{E}}{dx^{2}}(0) &= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} \frac{\mathcal{E}'(x)-\mathcal{E}'(0)}{x} \\
&= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} \frac{2}{x^{2}} \exp \left( -\frac{1}{x^2} \right) \\[2mm]
&= 0
\end{align*}$$
So, $\mathcal{E}''(0)=0$. I think you got the idea : you can prove (by induction, for example) that, for all $n \in \mathbb{N}^{\ast}$ there exists a polynomial $P_{n}$ such that 
$$
\begin{align*}
\frac{d^{n}\mathcal{E}}{dx^{n}}(0) &= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} \frac{\mathcal{E}^{(n-1)}(x)-\mathcal{E}^{(n-1)}(0)}{x} \\
&= \lim \limits_{\substack{x \to 0 \\ x \neq 0}} P_{n} \left( \frac{1}{x} \right) \exp \left( -\frac{1}{x^2} \right) \\[2mm]
&= 0
\end{align*}$$
A: Since $x \mapsto {1 \over x}$ is smooth for $x \neq 0$ and $x \mapsto e^x$ is smooth, it is clear that $\cal E$ is smooth for $x \neq 0$.
Suppose $x \ne 0$, then ${\cal E}^{(k)}$ has the form ${\cal E}^{(k)}(x) = e^{-{1 \over x^2}} p_k({1 \over x})$ for some polynomial $p_k$. This is clearly true for $k=0$, so suppose it is true for $k=0,...,n$. Then ${\cal E}^{(n)}(x) = e^{-{1 \over x^2}} p_n({1 \over x})$ and the chain rule gives
${\cal E}^{(n+1))}(x) = {\cal E}^{(1)}(x) p_n({1 \over x}) - {\cal E}^{(0)}(x) p_n'({1 \over x}) ({1 \over x^2}) = e^{-{1 \over x^2}} ({2 \over x^3}p_n({1 \over x})-p_n'({1 \over x}) ({1 \over x^2}) )$. If $p_{n+1}(y) = 2 y^3p_n(y)-p_n'(y) y^2 $, then ${\cal E}^{(n+1))}(x) = e^{-{1 \over x^2}} 
p_{n+1}( {1 \over x} ) $, and so the result is true for all $n$.
If $x \neq 0$, we have $e^{-{1 \over x^2}} = {1 \over {e^{1 \over x^2}}}$, and since $e^{1 \over x^2} \ge \sum_{k=0}^n {1 \over k!} {1 \over x^{2k}}$, we have
$e^{-{1 \over x^2}}  \le {x^{2n} \over \sum_{k=0}^n {1 \over k!} {x^{2(n-k)}}} \le {x^{2n} \over n!}$.
Suppose $p$ is a polynomial of degree $d$. Then for any $n$ we see that there is some constant $K$ such that $|e^{-{1 \over x^2}} p({1\over x})| \le K |x|^{2n-d}$ whenever $0 <|x| \le 1$. In particular, there is some $K$ such that
$|e^{-{1 \over x^2}} p({1\over x})| \le K x^2$ for all $0 < |x| \le 1$.
We have ${\cal E}^{(0)}(x) \le x^2$ for all $x$, and so ${\cal E}$ is continuous at $x=0$. Since
$|{\cal E}^{(0)}(x) - {\cal E}^{(0)}(0) -0| \le x^2$, we see that ${\cal E}^{(0)}$ is differentiable at $x=0$, and ${\cal E}^{(1)}(0) = 0$.
Now suppose ${\cal E}^{(k)}$ is differentiable at $x=0$ and ${\cal E}^{(k)}(0) = 0$ for $k=0,...,n$. Then
$|{\cal E}^{(n)}(x) - {\cal E}^{(n)}(0) -0| \le K x^2$ for some $K$ and $|x| \le 1$. Hence 
${\cal E}^{(n)}$ is differentiable at $x=0$, and ${\cal E}^{(n+1)}(0) = 0$.
A: Let $S=\left\{x\mapsto P\left(\frac{1}{x}\right)\mathcal E(x)\mid P\in \Bbb R\left[X\right]\right\}$
Can you prove that $f\in S \implies f'\in S$?
What can you say about $\lim\limits_{x\to 0}f(x)$ for $f \in S$?
