At a simple zero of $h$, dividing by $h$ produces a residue equal to numerator over $h'$.Suggest a slogan
Let \(g\) and \(h\) be holomorphic near \(z_0\), with \(h(z_0)=0\) and \(h'(z_0)\neq0\). Then \[ \Res_{z=z_0}\frac{g(z)}{h(z)}=\frac{g(z_0)}{h'(z_0)}. \] In particular, \(\Res_{z=z_0}\frac{1}{h(z)}=\frac{1}{h'(z_0)}\).