At absolute maximum or minimum derivative of #f(x)=(x+1)/(x^2+x+9)# will be zero. We have a maximum when second derivative is negative.

#(df)/(dx)=((x^2+x+9)xx1-(x+1)(2x+1))/(x^2+x+9)^2#

= #(x^2+x+9-2x^2-3x-1)/(x^2+x+9)^2#

= #-(x^2+2x-8)/(x^2+x+9)^2#

= #-((x+4)(x-2))/(x^2+x+9)^2#

as such we have extrema at #x=-4# and #x=2# and in the interval we have just one extrema at #x=2# and

#(d^2f)/(dx^2)=-((x^2+x+9)^2(2x+2)-2(x^2+x+9)(2x+1)(x^2+2x-8))/(x^2+x+9)^4#

= #-((x^2+x+9)((x^2+x+9)(2x+2)-(4x+2)(x^2+2x-8)))/(x^2+x+9)^4#

= #-(2x^3+4x^2+20x+18-(4x^3+10x^2-28x-16))/(x^2+x+9)^3#

= #(2x^3+6x^2-48x-34)/(x^2+x+9)^3#

at #x=2#, we have #(d^2f)/(dx^2)=-2/75#

and at #x=-4#, #(d^2f)/(dx^2)=2/147#

Hence we have an absolute maxima at #x=2# and #f(2)=3/15=1/5#

And as degree of denominator is higher, while at #x=0#, #f(0)=1/9#, as #x->oo#, #f(x)->0#

graph{y-(x+1)/(x^2+x+9)=0 [-10, 10, -0.4, 0.4]}

graph{y-(x+1)/(x^2+x+9)=0 [-200, 200, -0.4, 0.4]}