We can factor the numerator of the first fraction by finding its roots. In fact, every time an expression like [tex] ax^2+bx+c [/tex] has solutions [tex] x_1,x_2 [/tex], then the polynomial can be written as
[tex] ax^2+bx+c=a(x-x_1)(x-x_2) [/tex]
So, using the quadratic formula, i.e.
[tex] x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
we can find the solutions
[tex] x_{1,2} = \dfrac{1\pm\sqrt{1+24}}{2} = \dfrac{1\pm 5}{2}[/tex]
So, the two solutions are -2 and 3, which means that we have the following factorization:
[tex]x^2-x-6 = (x+2)(x-3) [/tex]
Moreover we can factor a 3 out of the denominator of that same fraction to get
[tex] 3x-9=3(x-3) [/tex]
So, the expression looks like this:
[tex] \dfrac{x^2-x-6}{3x-9}\cdot\dfrac{2}{x+2} = \dfrac{(x+2)(x-3)}{3(x-3)}\cdot\dfrac{2}{x+2}[/tex]
So, we can simplify a factor of x-3 from the first fraction, and cross-simplify a factor of x+2. We have
[tex] \dfrac{x+2}{3}\cdot\dfrac{2}{x+2} = \dfrac{2(x+2)}{3(x+2)} = \dfrac{2}{3}[/tex]
