I'm assuming your function is (x+3)/(x^2-9)
See attachment
No it is very possible though a little more difficult-- my impression is that at this level it would seem the one I suggested would be more relevant.
However for the original
| (x + 3)/(x-3)(x+3)| < |2x/(x+3)(x-3| since for x > 3 x + x > x+3
|2x/(x+3)(x-3)|< | 2x/(x)(x-3)| smaller denominator bigger fraction x < x+3
| 2x/(x)(x-3)| = 2/(x-3) x > 3
2/(x-3) < e
1/(x-3) < e/2
again if x- 3 > e/2
1/(x-3) < e/2
M = 3 + e/2