find x-int of
y=x^3-x^2-6x / -3x^2-3x+18
i know i have to make y=0, but im not sure where to go from there
yes, set y=0.
Since a fraction is zero when the numerator is zero, you just need
x^3-x^2-6x = 0
x(x-3)(x+2) = 0
But you have to check to see whether the denominator is zero at any of those values. If so, that would make y=0/0, which is undefined.
Luckily, your denominator is -3(x^2+x-6) = -3(x+3)(x-2)
so it is not zero at any of the roots of the numerator.
(unless you made a typo ...)