Find by any method the first three terms of Taylor expansion about $x=0$ of of $\exp\{-\frac{1}{(x-a)^2}\}$
I found derivatives and plugged them into the general form for Taylor series and got
$e^-\frac{1}{a^2}[1-\frac{2}{a^3}x+\frac{4-6a^2}{2a^6}x^2]$
My question is, why is it that we can sometimes use the more direct method of simply using the exponential power series (which then turns out to be the same as the Taylor series) as for example we can with the function $e^{-x}$, whereas if we try to do that here we get a series that looks very different?
Thinking about this more I'm guessing the answer has to do with implicit expansions appearing within the first expansion.
Can the Taylor series expansion of certain functions ever be substantially different to other expansions of the same function, or just different in form? What would be most appropriate form in the given question?