Here are my thoughts:
I think $z = 0$ is essential singularity so next step I need to write $e^{\left(\frac{1}{z^2}\right)}$ into the form of Laurent so I could find the value of b1, which is the residual.
And since essential singularity is different from the simple pole, I should not use the theorem of isolated singularity, right?
You can write$$e^{1/z^2}=1+\frac1{z^2}+\frac1{2z^4}+\frac1{3!z^6}+\cdots$$and it follows from this that $\operatorname{res}_{z=0}\left(e^{1/z^2}\right)=0$. But it is even simpler to note that, since it's an even function, its Laurent series $\sum_{n=-\infty}^\infty a_nz^n$ near $0$ must be such that $a_n=0$ when $n$ is odd. Therefore$$\operatorname{res}_{z=0}\left(e^{1/z^2}\right)=a_{-1}=0$$