How do you find the x-coordinate of any hole (s) in the graph of f(x)= x2-2x/x2-4?

Did you notice that

f(x)= x2-2x/x2-4
= x(x-2)/[(x+2)(x-2)]
= x/(x+2) , where x is not equal to 2

when you sub x=2 into the original, you get 0/0 which is indeterminate, but if you sub x=2 into the final result you get f(2) = 2/4 = 1/2

so there is a "hole" at (2,1/2)
and an asymtote at x=-2