Answer
$$F'(z)=-\frac{1}{z^2}(\frac{2A}{z}+B)$$
Work Step by Step
$$F(z)=\frac{A+Bz+Cz^2}{z^2}$$
$$F(z)=\frac{A}{z^2}+\frac{B}{z}+C$$
$$F(z)=Az^{-2}+Bz^{-1}+C$$
Therefore, the derivative of $F(z)$ is $$F'(z)=A\frac{d}{dz}(z^{-2})+B\frac{d}{dz}(z^{-1})+\frac{d}{dz}(C)$$
$$F'(z)=A\times(-2)z^{-3}+B\times(-1)z^{-2}+0$$
$$F'(z)=-2Az^{-3}-Bz^{-2}$$
$$F'(z)=-z^{-2}(2Az^{-1}+B)$$
$$F'(z)=-\frac{1}{z^2}(\frac{2A}{z}+B)$$
