Answer:
 (a)  There are asymptotes at x=3/2 and x=-1/3
Step-by-step explanation:
The denominator zeros can be found by factoring:
 f(x) = (x +1)/((2x -3)(3x +1))
Neither of the denominator factors is cancelled by the numerator factor, so each represents a vertical asyptote, not a function hole.
The asymptotes are at the values of x where the denominator is zero:
 2x -3 = 0  ⇒  x = 3/2
 3x +1 = 0  ⇒  x = -1/3