g(z) and h(z) are analytic, g'(z0)=h''(z0)not=0 but h(z0)=h'(z0)=0 show that the ration function g(z)/h(z) has a pole of order 2 and the residue is

(2g'/h'')-(2gh'''/3(h'')^2)=b_-1

g(z)=a0+a1(z-z0)+a2(z-z0)^2+....

1/h(z)=(b_-1/(z-z0)^2)+b0/(z-z0)+b1+....

g/h=(a0*b_-1/(z-z0)^2)+a1*b0+a2b1(z-z0)^2+....

so the res is at a0*b_-1?? if this is ok, then from here i have no idea... i must have to differentiate something 3 times but idk, can anyone plz help?