You can do the $\displaystyle 3z^2$ from the definition right? I'll do the other part:
Let $\displaystyle g(z)=i/z$
then:
$\displaystyle g'(z)=\lim_{\Delta z\to 0} \frac{\frac{i}{z+\Delta z}-\frac{i}{z}}{\Delta z}=\lim_{\Delta z\to 0}\frac{-i\Delta z}{z(z+\Delta z)\Delta z}=\lim_{\Delta z\to 0} \frac{-i}{z(z+\Delta z)}=-\frac{i}{z^2}$