Just to add some additional details to Alex M.'s response...this question can also be found in Spivak's Calculus, in Chapter 18 problem 40 (Ed. 4). There is an important lemma that is derived earlier in the exposition of the chapter, which reads as follows:
Theorem 6:For any natural number $n$: $\displaystyle \lim_{x \to \infty} \frac{e^x}{x^n}=\infty \quad (*_1)$.
We will need this for our proof:
Firstly, we will assume that we have already shown that for any $x \neq 0$ and $\displaystyle k \in \mathbb N: f^{(k)}(x)=e^{-\frac{1}{x^2}}\times\left(\sum_{i=1}^{3k}\frac{a_i}{x^i}\right) \quad (\dagger_1)$, where the $i$ used in $a_i$ is just meant as an index, but the $i$ used with $x^i$ refers to $x$ raised to some natural number power $i$. Note that many values of $a_i$ will be equal to $0$. Observe that $\sum_{i=1}^{3k}\frac{a_i}{x^i}$ is just some polynomial $P$ with $(\frac{1}{x})$ as its argument.
Although sort of clunky, this can be proven by induction in a straightforward manner by just taking the derivative of the above expression and showing that $\displaystyle f^{(k+1)}(x)=e^{-\frac{1}{x^2}}\times\left(\sum_{i=1}^{3(k+1)}\frac{b_i}{x^i}\right)$.
We will now show that for any arbitrary $k$, $f^{(k)}(0)=0$. We will also use induction to prove this. As the base case, recalling the definition of the derivative, we know that $f^{(1)}(0)=\displaystyle \lim_{h \to 0}\frac{f(h)-f(0)}{h}= \lim_{h \to 0}\frac{f(h)}{h}=\cdots \text{refer to the OPs work}\cdots=0$.
Next, assume that for an arbitrary $n$, we know that $f^{(n)}(0)=0$. We will now prove that $f^{(n+1)}(0)=0:$
By definition and $(\dagger_1)$, $f^{(n+1)}(0)=\displaystyle \lim_{h \to 0}\frac{f^{(n)}(h)-f^{(n)}(0)}{h}=\lim_{h \to 0}\frac{e^{-\frac{1}{h^2}}\times\left(\sum_{i=1}^{3k}\frac{a_i}{h^i}\right)-0}{h}$.
How exactly should we deal with the numerator? Well, $e^{-\frac{1}{h^2}}$ will be multiplied to every member of the summation...so let us just consider the $j$th arbitrary member of this expression: $e^{-\frac{1}{h^2}}\times \frac{a_j}{h^j}$. If we can show that $\displaystyle \lim_{h \to 0}\frac{e^{-\frac{1}{h^2}}\times \frac{a_j}{h^j}}{h}=0$, then the additive law of limits will let us conclude that $\displaystyle\lim_{h \to 0}\frac{e^{-\frac{1}{h^2}}\times\left(\sum_{i=1}^{3k}\frac{a_i}{h^i}\right)}{h}=0$.
It will be useful to rewrite $\displaystyle \lim_{h \to 0}\frac{e^{-\frac{1}{h^2}}\times \frac{a_j}{h^j}}{h}$ in the following way:
$$\frac{e^{-\frac{1}{h^2}}\times \frac{a_j}{h^j}}{h}=\frac{\frac{a_j}{h^{j+1}}}{e^{\frac{1}{h^2}}}=a_j \times \left(\frac{\frac{1}{h^{j+1}}}{e^{\frac{1}{h^2}}}\right)$$
To address this, let us prove the more general statement:
For any $\displaystyle n \in \mathbb N: \lim_{h \to 0}\frac{\frac{1}{h^n}}{e^{\frac{1}{h^2}}}=0$
To prove this, we will break down the above statement into two different cases:
$\displaystyle \lim_{h \to 0^+}\frac{\frac{1}{h^n}}{e^{\frac{1}{h^2}}}=0$
$\displaystyle \lim_{h \to 0^-}\frac{\frac{1}{h^n}}{e^{\frac{1}{h^2}}}=0$
Note that limit laws let us evaluate the following two equivalent cases:
$\displaystyle \lim_{h \to \infty}\frac{h^n}{e^{h^2}}=0$
$\displaystyle \lim_{h \to -\infty}\frac{h^n}{e^{h^2}}=0$
Recall from $(*_1)$ that $\displaystyle \lim_{x \to \infty}\frac{e^x}{x^n}=\infty$. Because $\exp$ is a strictly increasing function, for $x \gt 1$, we note that $e^{x} \lt e^{x^2}$. Therefore, we most certainly have that $\displaystyle \lim_{x \to \infty}\frac{e^x}{x^n}=\infty \implies \displaystyle \lim_{x \to \infty}\frac{e^{x^2}}{x^n}=\infty$. Of course, then, the inverse of such an expression must equal $0$...i.e $\displaystyle \lim_{x \to \infty}\frac{e^{x^2}}{x^n}=\infty \implies \displaystyle \lim_{x \to \infty}\frac{x^n}{e^{x^2}}=0$. This deals with case 1.
Because $e^{x^2}$ is an even function, the relevant consideration for case 2 is whether or not $n$ is even or odd. If $n$ is even, then $\displaystyle \lim_{x \to -\infty}\frac{x^n}{e^{x^2}} = \lim_{x \to \infty}\frac{x^n}{e^{x^2}}=0$. But if $n$ is odd, then $\displaystyle \lim_{x \to -\infty}\frac{x^n}{e^{x^2}}=-\displaystyle \lim_{x \to \infty}\frac{x^n}{e^{x^2}}=-1 \times 0=0$. With the two cases established, we conclude that for any $\displaystyle n \in \mathbb N: \lim_{h \to 0}\frac{\frac{1}{h^n}}{e^{\frac{1}{h^2}}}=0$.
Of course, then, we have that $\displaystyle \lim_{h \to 0} a_j \times \left(\frac{\frac{1}{h^{j+1}}}{e^{\frac{1}{h^2}}}\right)=a_j \lim_{h \to 0}\left(\frac{\frac{1}{h^{j+1}}}{e^{\frac{1}{h^2}}}\right)=a_j \times 0=0$
