Answer
$\displaystyle \frac{2x^{2}-3}{ (x-3)(x+3)(x-2)}, \qquad x\neq\pm 3,2$
Work Step by Step
We find the common denominator first.
$ x^{2}-9=(x+3)(x-3)\qquad $(a difference of squares)
$ x^{2}-5x+6=(x-2)(x-3)$
($-2$ and $-3$ are factors of $+6$ whose sum is $-5$)
We list the common factors first:
$LCD=(x-3)(x+3)(x-2)$
Domain: $ x\neq\pm 3,2$
Multiply each term with 1 (either $\displaystyle \frac{x-3}{x-3}, \displaystyle \frac{x+3}{x+3}$ or $\displaystyle \frac{x-2}{x-2}$)
$\displaystyle \frac{x}{ (x-3)(x+3)}\cdot\frac{x-2}{x-2}+\frac{x-1}{(x-2)(x-3)}\cdot\frac{x+3}{x+3}=$
$=\displaystyle \frac{x(x-2)+(x-1)(x+3)}{ (x-3)(x+3)(x-2)}$
$=\displaystyle \frac{x^{2}-2x+x^{2}+3x-x-3}{ (x-3)(x+3)(x-2)}$
$=\displaystyle \frac{2x^{2}-3}{ (x-3)(x+3)(x-2)}, \qquad x\neq\pm 3,2$
