Simplify the inequality, we cannot do crossing over.
x/(x-2) > -1/(x+3)
x/(x-2)+1/(x+3)>0
(x(x+3)+(x-2))/((x-2)(x+3))>0
(x^2+3x+x-2)/((x-2)(x+3)) >0
(x^2+4x-2)/((x-2)(x+3)) >0
The roots of the numerator
x^2+4x-2=0
are
x=(-4+-sqrt(16-4(1)(-2)))/(2)
=-2+-sqrt6
x_1=-2-sqrt6=-4.45
x_2=-2+sqrt6=0.45
Let
f(x)=((x-x_1)(x-x_2))/((x-2)(x+3))
We can build the sign chart
color(white)(aaaa)xcolor(white)(aaaa)-oocolor(white)(aaaa)x_1color(white)(aaaa)-3color(white)(aaaaa)x_2color(white)(aaaaa)2color(white)(aaaaa)+oo
color(white)(aaaa)x-x_1color(white)(aaaa)-color(white)(aaaa)+color(white)(aaaa)+color(white)(aaaa)+color(white)(aaaa)+color(white)(aaaa)
color(white)(aaaa)x-3color(white)(aaaaa)-color(white)(aaaa)-color(white)(aa)||color(white)(a)+color(white)(aaaa)+color(white)(aaaa)+
color(white)(aaaa)x-x_2color(white)(aaaa)-color(white)(aaaa)-color(white)(aa)#color(white)(aa)-#color(white)(aaaa)+color(white)(aaaa)+
color(white)(aaaa)x-2color(white)(aaaaa)-color(white)(aaaa)-color(white)(aa)#color(white)(aa)-#color(white)(aaaa)-color(white)(aa)||color(white)(a)+
color(white)(aaaa)f(x)color(white)(aaaaaa)+color(white)(aaaa)-color(white)(aa)||color(white)(a)+color(white)(aaaa)-color(white)(aa)||color(white)(a)+
Therefore,
f(x) >0 when x in (-oo,-4.45] uu(-3, 0.45] uu (2, +oo)
graph{(x^2+4x-2)/((x-2)(x+3)) [-10, 10, -5, 5]}
