 Solution

Insert $u=-1/x^2$ into the exponential series $e^u={\sum }_{n=0}^{\infty }{u}^n/n!$. For every fixed $x\ne 0$, this gives the expansion with absolute convergenceThis is a Laurent series in inverse powers of $x$, as requested. It is not a Maclaurin series about zero: that series has every coefficient zero and cannot equal $f$ on a punctured neighborhood of zero. There is no contradiction, since the inverse-power expansion does not apply at $x=0$.