find the derivative of
g(y)= (y^2/(y+3))^3
use the quotient rule
g'(y) = [(y+3)^3(2y) - y^2(3)(y+3)^2]/(y+3)^6
= y(y+3)^2[2(y+3) - 3y]/(y+3)^6
= y(6-y)/(y+3)^4
g(y)= (y^2/(y+3))^3
g'(y) = [(y+3)^3(2y) - y^2(3)(y+3)^2]/(y+3)^6
= y(y+3)^2[2(y+3) - 3y]/(y+3)^6
= y(6-y)/(y+3)^4