Solution-
Equating the denominator to 0,
⇒ (x+3)(x+1) = 0
⇒ x= -3, -1
∴ Discontinuities of the function f(x) is at -3, -1
As (x+3) is also a factor of the numerator, you will have what's called a removable discontinuity. That will show up as a hole in the graph.
∴ At x = -3, you will have removable discontinuity.