One way of rewriting the expression is:
[√(x²+x-2)²]/(x²-x-6).
A more explicit way is to express it piecewise:
x²+x-2=(x-1)(x+2) and x²-x-6=(x-3)(x+2).
There are 3 critical points: x=-2, 1, 3 and the expression is discontinuous.
The common factor x+2 cancels to give 1 or -1 and gives rise to one of the discontinuities. The other discontinuity is at x=3.
Let f(x)=|x²+x-2|/(x²-x-6), then f can be defined piecewise:
f(x)=
⎛(1-x)/(3-x) | x<-2
⎥undefined | x=-2
⎥(1-x)/(x-3) | -2<x<1
⎥(x-1)/(x-3) | 1≤x<3
⎥undefined | x=3
⎝(x-1)/(x-3) | x>3