#f(x) = (x+3)/(x^2-9) = color(red)(cancel(color(black)(x+3)))/((x-3)color(red)(cancel(color(black)((x+3))))) = 1/(x-3)#

excluding #x=-3#

As #x->+-oo#, #1/(x-3)->0#. So there is a horizontal asymptote #y=0#

As #x->3^+#, #1/(x-3)->+oo#

As #x->3^-#, #1/(x-3)->-oo#

#f(3)# is undefined since division by #0# is undefined.

So #f(x)# has a vertical asymptote at #x=3#

#f(-3)# is undefined, since both numerator #(x+3) = 0# and denominator #(x^2-9) = 0#. Note however, that both left and right limits exist at #x=-3# and are both equal to #-1/6#. So there is a removable singularity at #x=-3# (removable by redefining #f(-3) = -1/6#).