At a simple zero of the denominator, the residue is numerator over derivative: $g(z_0)/h'(z_0)$.Suggest a slogan
Let \(g\) and \(h\) be holomorphic near \(z_0\), with \(h(z_0)=0\) and \(h'(z_0)\neq0\). Then \[ \Res_{z=z_0}\frac{g(z)}{h(z)}=\frac{g(z_0)}{h'(z_0)}. \]
Write \(h(z)=(z-z_0)h_1(z)\) with \(h_1\) holomorphic near \(z_0\) and \(h_1(z_0)=h'(z_0)\neq0\). Then \(g/h\) has at most a simple pole at \(z_0\), and its residue is \(\lim_{z\to z_0}(z-z_0)\frac{g(z)}{h(z)}=\frac{g(z_0)}{h_1(z_0)}=\frac{g(z_0)}{h'(z_0)}\).