Answer
$$ g'(z)=\frac{-\csc^2z+\sin z\csc^2z+\cot z \cos z}{3(1-\sin z)^2}$$
Work Step by Step
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Recall that $(\cot x)'=-\csc^2 x$.
Recall that $(\sin x)'=\cos x$.
Since $ g(z)=\frac{\cot z}{3-3\sin z}$, then using the quotient rule, we have
$$ g'(z)=\frac{(3-3\sin z)(\cot z)'-\cot z (3-3\sin z)'}{(3-3\sin z)^2}
\\=\frac{(3-3\sin z)(-\csc^2z)-\cot z (-3\cos z)}{(3-3\sin z)^2}=\frac{-3\csc^2z+3\sin z\csc^2z+3\cot z \cos z}{(3-3\sin z)^2}\\
=\frac{3(-\csc^2z+\sin z\csc^2z+\cot z \cos z)}{9(1-\sin z)^2}\\
=\frac{-\csc^2z+\sin z\csc^2z+\cot z \cos z}{3(1-\sin z)^2}
$$
