I think you've made a mistake math applicant. The pole of e^(z^2)-1 at z=0 is order 2. Notice that the first derivative is 2ze^(z^2) which has a zero at z=0 and taking another derivative we find that the second derivative is not zero, so the zero is second order. Ergo e^(z^2)-1/z^2 is entire. Anyway I'd suggest taking the Taylor series for e^(z^2)-1 and then dividing by z^2 to get the correct Taylor series which should give you all the derivatives you could possibly need.