Let #f(x)=x^2/(x^2+9)#
#f(-x)=x^2/(x^2+9)#
#f(x)=f(-x)#
The curve is symmetric about the y-axis
The derivative of a quotient is
#(u/v)'=(u'v-uv')/(v^2)#
We start by calculating the first derivative
#y=x^2/(x^2+9)#
#u=x^2#, #=>#, #u'=2x#
#v=x^2+9#, #=>#, #v'=2x#
#dy/dx=(2x(x^2+9)-2x(x^2))/(x^2+9)^2#
#=(2x^3+18x-2x^3)/(x^2+9)^2#
#=(18x)/(x^2+9)#
The critical values are when #dy/dx=0#
#(18x)/(x^2+9)^2=0#
When #x=0#
We can build a chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##0##color(white)(aaaaaa)##+oo#
#color(white)(aaaa)##dy/dx##color(white)(aaaaaa)##-##color(white)(aaa)##0##color(white)(aaaa)##+#
#color(white)(aaaa)##y##color(white)(aaaaaaaa)##↘##color(white)(aaa)##0##color(white)(aaaa)##↗#
Now, we calculate the second derivative
#u=18x#, #=>#, #u'=18#
#v=(x^2+9)^2#, #=>#, #v'=2(x^2+9)*2x#
#(d^2y)/dx^2=(18(x^2+9)^2-18x*(4x(x^2+9)))/(x^2+9)^4#
#=18(((x^2+9)^2-4x^2(x^2+9)))/(x^2+9)^4#
#=(18(x^2+9)((x^2+9)-4x^2))/(x^2+9)^4#
#=(18(9-3x^2))/(x^2+9)^3#
#=(54(3-x^2))/(x^2+9)^3#
#(d^2y)/dx^2=0# when #x =-sqrt3# and #x=sqrt3#
The points of inflexions are #(-sqrt3, 1/4)# and #(sqrt3,1/4)#
We can build the chart
#color(white)(aa)##Interval##color(white)(aaaa)##]-oo,-sqrt3[##color(white)(aaaa)##]-sqrt3,sqrt3[##color(white)(aaaa)##]sqrt3,+oo[#
#color(white)(aa)##(d^2y)/dx^2##color(white)(aaaaaaaaaaaaa)##-##color(white)(aaaaaaaaaaaa)##+##color(white)(aaaaaaaaa)##-#
#color(white)(aa)##y##color(white)(aaaaaaaaaaaaaaaa)##nn##color(white)(aaaaaaaaaaaa)##uu##color(white)(aaaaaaaaa)##nn#
#lim_(x->+-oo)y=lim_(x->+-oo)x^2/x^2=1#
The horizontal asymptote is #y=1#
graph{(y-(x^2)/(x^2+9))(y-1)=0 [-7.02, 7.024, -3.51, 3.51]}