y'= (x^3+6) * (d/dx)(x^2+x-2) - (x^2+x-2) *(d/dx)(x^3+6)/(x^3+6)^2

You take the derivative of the one behind the d/dx so you get:

(x^3+6)(2x+1) - (x^2+x-2)(3x^2)/(x^2+6)^2

so in other words you take the derivative of (x^2+x-2) to get (2x+1) and you take the derivative of (x^3+6) to get (3x^2)

I hope this helps