#36/[x^2-9]= (2x)/[x + 3]- 1/[1]#

#1/1# can be anything, make the right side denominators equal.

#36/[color(red)(x^2-9)]= color(blue)((2x)/[x + 3]- (x+3)/(x+3))" "larr# factorise and simplfy

#36/color(red)[(x+3)(x-3)]=color(blue)( (2x -(x+3))/(x+3))#

#36/[(x+3)(x-3)]= (2x -x-3)/(x+3)#

#36/[(x+3)(x-3)]= (x-3)/(x+3)" "color(forestgreen)(xx (x+3))#

#(36color(forestgreen)(xx cancel((x+3))))/[cancel((x+3))(x-3)]= ((x-3)color(forestgreen)(xxcancel ((x+3))))/cancel((x+3))" "larr# cancel

#36/((x-3)) = (x-3)" "larr# cross multiply

#(x-3)^2 =36" "larr# square root both sides

#x- 3= +-6#

If #x-3 =+ 6" "rarr x = 9#

If #x-3=-6" "rarr x = -3#

However, #x=-3# will make the denominators 0, so this is an extraneous solution which we will reject.