Observe that if #x=+-a#, #y=a^2/(1+a^2)# i.e. for every #(x,y)#, #(-x,y)# too lies on the curve. Hence, the graph is symmetric w.r.t. #y#-axis.
Now as #y=x^2/(1+x^2)=(1+x^2-1)/(1+x^2)=1-1/(1+x)^2=1-(1+x^2)^(-2)# and hence,
#(dy)/(dx)=-(-2/(1+x^2)^3xx2x)=(4x)/(1+x^2)^3#
Observe that as #y=1-1/(1+x)^2#, value of #y# or range of #x^2/(1+x^2)# is limited between #[0,1)#. Further, as #(dy)/(dx)=0# only at #x=0# and #x=+-oo# and hence we have an extrema at these points.
Now for second derivative using quotient formula, it is
#(d^2y)/(dx^2)=((1+x^2)^3xx4-4x xx3(1+x^2)^2xx2x)/(1+x^2)^6#
= #((1+x^2)^2[4+4x^2-24x^2))/(1+x^2)^6=(4(1-5x^2))/(1+x^2)^4#
and this is zero when #x^2=0.2# or #x=+-0.447#
Alsso note that when #x=0#, #(d^2y)/(dx^2) > 0#, hence wehave a minima at #x=0# and when #x->+-oo#, #(d^2y)/(dx^2) -> 0#, but on the negative side. Further, as #x->+-oo#, #y->1# and hence we have a maxima at #+-oo#.
The graph appears as follows:
graph{x^2/(1+x^2) [-5, 5, -1.62, 3.38]}
