In vicinity of the point x= -2,  


    x^2 - x - 6 = (x-3)*(x+2),


so, after dividing by (x+2)


    f(x) =  = x-3  at  x < -2.


So, the limit of f(x) at x---> -2 from the left  is equal to -5.


From the other side,  f(-2) = -5, by the definition of the function f(x)  (second line).


Also, the limit f(x) at x ---> -2 from the right is equal to -5.


So, the function f(x) is CONTINUOUS at x= -2.




Next, at x= 3, the function f(x) is not formally defined, so we even can not discuss the question if it is continuous at this point.


It is NOT.

    The way how the function f(x) is defined in the third line, 


        f(x) = { (x^2-9)/(x-3)  , if x≥3


    is a  HUGE mistake.   It is impossible to define a function in this way, since division by  0  at x= 3 is STRICTLY PROHIBITED.


    The correct form of this third line should be THIS


        f(x) = { (x^2-9)/(x-3)  , if x > 3.


    At x = 3, the function f(x) is UNDEFINED.