Answer:
the removable discontinuities are at x = 6 and at x = -6
Explanation:
Notice that the function has common binomial factor in numerator and denominator:

therefore, the removable discontinuities are those at x= 6 and x = -6 that correspond to zeros common in numerator and denominator, and therefore those associated with the (x + 6) factor, with the (x - 6) factor.
There is a non-removable discontinuity at x = 0.
The discontinuities can be removed by re-assigning the value of f(x) at x=6 (as 1/6), and at x = -6 (as -1/6)