The function is
#y=x/(x^2-9)#
The domain is #x in (-oo,-3) uu(-3,3) uu(3,+oo)#
The first derivative is calculated with the quotient rule
#(u/v)'=(u'v-uv')/(v^2)#
Here,
#u=x#, #=>#, #u'=1#
#v=x^2-9#, #=>#, #v'=2x#
#y'=dy/dx=(1*(x^2-9)-(x*2x))/(x^2-9)^2#
#=(x^2-9-2x^2)/(x^2-9)^2#
#=(-9-x^2)/(x^2-9)^2#
There are no critical points
#y'!=0#, #AA x in D_y#
Let's calculate the second derivative with the quotient rule
#u=-9-x^2#, #=>#, #u'=-2x#
#v=(x^2-9)^2#, #=>#, #v'=2(x^2-9)*2x#
Therefore,
#y''=(d^2y)/dx^2=(-2x(x^2-9)^2-(-9-x^2)(4x(x^2-9)))/(x^2-9)^4#
#=(-2x^3+18x+36x+4x^3)/(x^2-9)^3#
#=(2x^3+54x)/(x^2-9)^3#
#=(2x(x^2+27))/(x^2-9)^3#
When #y''=0#, there is a point of inflection.
#(2x(x^2+27))/(x^2-9)^3=0#
#=>#, #x=0#
Let's make a variation chart
#color(white)(a)##" Interval "##color(white)(aaa)##(-oo, -3)##color(white)(aaa)##(-3,0)##color(white)(aaa)##(0,3)##color(white)(aaa)##(3,+oo)#
#color(white)(a)##" Sign y'' "##color(white)(aaaaaaaa)##-##color(white)(aaaaaaaaa)##+##color(white)(aaaaaa)##-##color(white)(aaaaaa)##+#
#color(white)(a)##" y "##color(white)(aaaaaaaaaaaaa)##nn##color(white)(aaaaaaaaaa)##uu##color(white)(aaaaaa)##nn##color(white)(aaaaaa)##uu#
graph{x/(x^2-9) [-10, 10, -5, 5]}
