= Solution
Factor the denominator as $(z-z_0)(z+z_0)$. Since $z_0\ne0$, its zeros are distinct, and the potential simple <poles> and their <residues> are
$$
\boxed{\operatorname{Res}_{z=z_0}f=\frac{g(z_0)}{2z_0},\qquad
\operatorname{Res}_{z=-z_0}f=-\frac{g(-z_0)}{2z_0}.}
$$
For example, the first follows by taking $\lim_{z\to z_0}(z-z_0)f(z)$. If the numerator vanishes at either denominator zero, the singularity there is removable and the corresponding <residue> is zero. Thus the data identify the possible <poles>, not an unconditional <pole> at a point canceled by $g$.
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