Why doesn't the Taylor expansion at 0 around $ e^{-1/x^2} $ converge to the function itself?

The Taylor expansion does converge - to the function that's everywhere $0$. For every $n$ the remainder term at $x$ (the error in the Taylor expansion) is $\exp(-1/x^2)$, the value of the function you started with.

Since the Maclaurin polynomial is zero for each integer $n$, the remainder term is just $f$ itself, and since this does not vanish for $x\neq 0,\ f$ is not represented by a Taylor series ar $x=0.$