x²+x-6=(x+3)(x-2), so (x²+x-6)/(x-2)=x+3 for all values of x except x=2. So there is a discontinuity at x=2 only. Note that when x=-3, we get 0/(-5)=0 making the function continuous at x=-3. The discontinuity at x=2 is because we get 0/0 which cannot be evaluated, but it is just a singularity (hole) that does not occur for any other value of x.