The function is
#f(x)=(x+3x^2)/(1-x^2)#
This function is a quotient of #2# derivable functions
The derivative of a quotient is
#(u/v)'=(u'v-uv')/(v^2)#
#u=x+3x^2#, #=>#, #u'=1+6x#
#v=1-x^2#, #=>#, #u'=-2x#
Therefore,
#f'(x)=((1+6x)(1-x^2)-(x+3x^2)(-2x))/(1-x^2)^2#
#=(1-x^2+6x-6x^3+2x^2+6x^3)/(1-x^2)^2#
#=(x^2+6x+1)/(1-x^2)^2#
The critical points are when
#f'(x)=0#
That is
#x^2+6x+1=0#
The solutions to this quadratic equation are
#x=(-6+-sqrt(36-4))/(2)=(-6+-sqrt(32))/(2)#
#=(-6+-4sqrt2)/(2)#
#=-3+-2sqrt2#
Let #x_1=-3-2sqrt2#
and
#x_2=-3+2sqrt2#
Let's build a variation chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##x_1##color(white)(aaaa)##-1##color(white)(aaaaaaa)##x_2##color(white)(aaaa)##1##color(white)(aaaa)##+oo#
#color(white)(aaaa)##f'(x)##color(white)(aaaa)##+##color(white)(aa)##0##color(white)(aaa)##-##color(white)(aaa)##-##color(white)(aaaa)##0##color(white)(aa)##+##color(white)(aa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaa)##↗##color(white)(aa)####color(white)(aaa)##↘##color(white)(aaa)##↘##color(white)(aaa)####color(white)(aa)##↗##color(white)(aaa)##↗#
Therefore,
There is a local maximum at #(-5.83, -2.91)# and a local minimum at #(-0.17, -0.86)#.
graph{(x+3x^2)/(1-x^2) [-10, 10, -5, 5]}