Trying to find an explicit formula for the $n$th derivative of a function.

Question

So my function is $$ f: \mathbb{R}\setminus \{0\}\to \mathbb{R},\; f(x)=\exp\Bigl(\frac{-1}{x^2}\Bigr). $$ My task is to find a formula for the $n$-th derivative of this function. When I wrote out the first five derivatives: \begin{align} f'(x) &= \frac{2\exp(\frac{-1}{x^2})}{x^3}, \\ f''(x) &= \frac{-2\exp(\frac{-1}{x^2})(3x^2-2)}{x^6}, \\ f'''(x) &= \frac{4\exp(\frac{-1}{x^2})(6x^4-9x^2+2)}{x^9}, \\ f''''(x) &= \frac{-4\exp(\frac{-1}{x^2}) (30x^6-75x^4+36x^2-4)}{x^{12}}, \\ f'''''(x) &= \frac{8\exp(\frac{-1}{x^2}) (90x^8-330x^6+225x^4-60x^2+4)}{x^{15}} \end{align} I noticed some patterns. The explicit formula has to look somewhat like this: $$ f^{(n)}(x) = \frac{(\text{coefficient}) \cdot \exp(\frac{-1}{x^2}) \cdot (\text{polynomial})}{x^{3n}}. $$ I noticed that the coefficient alternates (for uneven $n$ positive and for even $n$ negative) so I can write this with something like $(-1)^{n+1}$ also the coefficient is always in the form of $2^k, k\in\mathbb{N}$. And it's similar to the last number in the polynomial. I just cant seem to find the whole pattern for the polynomial. Can anyone help?