By the Quotient Rule, the derivative is
#f'(x)=((x^2+1)(2x+1) - (x^2+x)(2x))/((x^2+1)^2)#
#=(2x^3+x^2+2x+1-2x^3-2x^2)/((x^2+1)^2)#
#=(-x^2+2x+1)/((x^2+1)^2)#
The second derivative is (also using the Chain Rule),
#f''(x)#
#=((x^2+1)^2*(-2x+2)-(-x^2+2x+1)*2(x^2+1)*2x)/((x^2+1)^4)#
#=((x^2+1)*(-2x+2)-(-x^2+2x+1)*4x)/((x^2+1)^3)#
#=(-2x^3+2x^2-2x+2+4x^3-8x^2-4x)/((x^2+1)^3)#
#=(2x^3-6x^2-6x+2)/((x^2+1)^3)#.
Inflection points occur where #f''(x)# changes sign (also note that #f''(x)# is defined for all #x#). The #x#-values where #f''(x)=0# occur where #2x^3-6x^2-6x+2=0 \Leftrightarrow x^3-3x^2-3x+1=0#.
You can either use your calculator or the Rational Root Theorem to help you find that #x=-1# is a root of this (once you know it's a root, it's easy to check: #(-1)^3-3(-1)^2-3(-1)+1=-1-3+3+1=0#.)
Next, either use long division of polynomials or synthetic division to see that #x^3-3x^2-3x+1=(x+1)(x^2-4x+1)#.
The quadratic formula gives the other two roots of #f''(x)# as
#x=(4 pm sqrt{16-4})/(2)=2 pm 1/2 *sqrt{12}=2 pm sqrt{3}#
You can most easily see that #f''(x)# changes sign at these three values by graphing it. But, since #f''(x)# is continuous for all #x#, you can just check its sign on various intervals by plugging in sample points.
Therefore, the #x#-coordinates of the three inflection points of #f(x)# are #x=-1, 2 pm sqrt(3)#.
By plugging these numbers into #f(x)# and doing various manipulations, you can find that the full rectangular coordinates of the three inflection points are: #(-1,0)#, #(2+sqrt{3},3/4+sqrt{3}/4)#, and #(2-sqrt{3},3/4-sqrt{3}/4)#.
