We have to find the derivative of  y = (3x^3+3)/(3x^3-3)

We use the quotient rule which state that the value of [f(x) / g(x)]' = [f'(x)*g(x) - f(x)*g'(x)]/ [g(x)]^2

Substituting the functions we have:

[(3x^3+3)/(3x^3-3)]' = [(3x^3+3)'*(3x^3-3) - (3x^3+3)*(3x^3-3)']/ [(3x^3-3)]^2

=> [9x^2*(3x^3-3) - (3x^3+3)*9x^2]/ [(3x^3-3)]^2

=> [27x^5 - 27x^2 - 27x^5 - 27x^2] / [(3x^3-3)]^2

=> [- 27x^2 - 27x^2] / [(3x^3-3)]^2

=> [- 54x^2 ] / [(3x^3-3)]^2

=> [- 54x^2 ] / 9*[(x^3-1)]^2

=> [-6x^2 ] /[(x^3-1)]^2

The required result is [-6x^2 ] /[(x^3-1)]^2