We cannot do crossing over, so rewrite the inequality
#3/(x-2)<=3/(x+3)#
#3/(x-2)-3/(x+3)<=0#
Placing on the same denominator
#(3(x+3)-3(x-2))/((x-2)(x+3))<=0#
#(3x+9-3x+6)/((x-2)(x+3))<=0#
#15/((x-2)(x+3))<=0#
Let #f(x)=15/((x-2)(x+3))#
The domain of #f(x)# is #D_f(x)=RR-{-3,2}#
Now, we can build the sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-3##color(white)(aaaaaa)##2##color(white)(aaaaaaa)##+oo#
#color(white)(aaaa)##x+3##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aa)##||##color(white)(aaaa)##+#
#color(white)(aaaa)##x-2##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aa)##||##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaa)##+##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aa)##||##color(white)(aaaa)##+#
Therefore,
#f(x)<=0# when #x in ]-3,2[#