Answer
$G'(y)=-\dfrac{3ABy^2}{(Ay^3+B)^2}$
Work Step by Step
$G(y)=\dfrac{B}{Ay^3+B}$
$A$ and $B$ are constants while $y$ is the variable. This is important to know because the derivative of a constant is always 0.
Use the quotient rule:
$\Bigg(\dfrac{f}{g}\Bigg)'=\dfrac{gf'-fg'}{g^2}$
so...
$G'(y)=\dfrac{(Ay^3+B)(0)-(B)(3Ay^2+0)}{(Ay^3+B)^2}$
Perform multiplication:
$G'(y)=\dfrac{-3ABy^2}{(Ay^3+B)^2}$'
which can be rewritten as:
$G'(y)=-\dfrac{3ABy^2}{(Ay^3+B)^2}$
