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]}
