Find the power series representation of $e^{-x^2}$ I know that the Maclaurin expansion of $e^x$ is $$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
But i'm not sure how to find the Maclaurin series here
I tried this
$$
f'_{(0)}=-2xe^{-x^2}=0
$$
And that follows to every derivative that follows, so how can I get a power series out of it?

That is easier than what you think.
General expansion:
$$e^A = 1 + A + \frac{1}{2}A^2 + \frac{1}{6}A^3 + \cdots$$
So if $A = -x^2$ you get
$$e^{-x^2} = 1 - x^2 + \frac{1}{2}(-x^2)^2 + \frac{1}{6}(-x^2)^3 + \cdots$$
So
$$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$$
To find the general series expansion, you just have to remember the series expansion of $e^A$:
$$e^A = \sum_{k = 0}^{+\infty} \frac{A^k}{k!}$$
So again: if $A = -x^2$ you gain finally
$$\boxed{e^{-x^2} = \sum_{k = 0}^{+\infty} \frac{(-x^2)^k}{k!} = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} x^{2k}}$$
