= Solution
Multiply the exponential <power series> by the two prefactors:
$$
f(z)=\sum_{n=0}^{\infty}\frac{z^{1-n}}{n!}+\sum_{n=0}^{\infty}\frac{z^{-1-n}}{n!}.
$$
Collecting powers yields the <Laurent series>
$$
\boxed{f(z)=z+1+\sum_{m=1}^{\infty}\left(\frac1{(m+1)!}+\frac1{(m-1)!}\right)z^{-m},\qquad0<|z|<\infty.}
$$
Its first terms are $z+1+3/(2z)+7/(6z^2)+13/(24z^3)+\cdots$. Infinitely many negative-power coefficients are nonzero, so $\boxed{z=0\text{ is an essential singularity}}$. The <residue> is the coefficient of $z^{-1}$:
$$
\boxed{\operatorname{Res}_{z=0}f=\frac1{2!}+\frac1{0!}=\frac32.}
$$
The phrase "power series" here refers to a <Laurent series>; no ordinary <Taylor series> about an <essential singularity> exists.
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