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)xcolor(white)(aaaa)-oocolor(white)(aaaaa)-3color(white)(aaaaaa)2color(white)(aaaaaaa)+oo
color(white)(aaaa)x+3color(white)(aaaa)-color(white)(aaaa)||color(white)(aa)+color(white)(aa)||color(white)(aaaa)+
color(white)(aaaa)x-2color(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[
