= Solution
Expand the exponential in inverse powers of $z$:
$$
g(z)=\sum_{n=0}^\infty\frac{z^{-n}}{n!}-\sum_{n=0}^\infty\frac{z^{2-n}}{n!}
=-z^2-z+\frac12+\sum_{k=1}^\infty\left(\frac1{k!}-\frac1{(k+2)!}\right)z^{-k}.
$$
This is a <Laurent series>, not a <Taylor series>, and converges throughout $0<|z|<\infty$. Every displayed coefficient of a negative power is nonzero, so its principal part has infinitely many terms. Thus
$$
\boxed{z=0\ \text{is an essential singularity}}.
$$
The <residue> is the coefficient of $z^{-1}$:
$$
\boxed{\operatorname{Res}_{z=0}g=1-\frac1{3!}=\frac56}.
$$
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