Answer
$\frac{3y(y^2+2-6y^3)}{(y^2+2)^3}$
Work Step by Step
Start with cancelling the common factor $(y^2+2)^4$, we have: $\frac{(y^2+2)^5(3y)-y^3(6)(y^2+2)^4(3y)}{(y^2+2)^7}=\frac{(y^2+2)(3y)-y^3(6)(3y)}{(y^2+2)^3}
=\frac{3y(y^2+2)-18y^4}{(y^2+2)^3}
=\frac{3y(y^2+2-6y^3)}{(y^2+2)^3}$
