We need

#(u/v)'=(u'v-uv')/(v^2)#

Our function is

#y=(x^5-4x^3+3x)/(x^2-1)#

#u(x)=x^5-4x^3+3x#, #=>#, #u'(x)=5x^4-12x^2+3#

#v(x)=x^2-1#, #=>#, #v'(x)=2x#

Therefore,

#dy/dx=((5x^4-12x^2+3)(x^2-1)-(x^5-4x^3+3x)(2x))/(x^2-1)^2#

#=((5x^4-12x^2+3)(x^2-1)-x(x^2-1)(x^2-3)(2x))/(x^2-1)^2#

#=((x^2-1)(5x^4-12x^2+3-2x^4+6x^2))/(x^2-1)^2#

#=(3x^4-6x^2+3)/(x^2-1)#

#=(3(x^4-2x^2+1))/(x^2-1)#

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

#=3(x^2-1)#

The critical points are when #dy/dx=0#

#3(x^2-1)=3(x+1)(x-1)=0#, #=>#, #x=1# and #x=-1#

We can build the variation chart

#color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,-1)##color(white)(aaaa)##(-1,1)##color(white)(aaaa)##(1,+oo)#

#color(white)(aaaa)##Sign dy/dx##color(white)(aaaaaaaaa)##+##color(white)(aaaaaaaaaa)##-##color(white)(aaaaaaaa)##+#

#color(white)(aaaa)##y##color(white)(aaaaaaaaaaaaaaa)##↗##color(white)(aaaaaaaaaa)##↘##color(white)(aaaaaaaa)##↗#

The intervals of increasing are # x in (-oo,-1) uu (1,+oo)#

The interval of decreasing is #x in (-1,1)#

We could have obtained the same result by simplifying #y#

#y=(x^5--4x^3+3x)/(x^2-1)=(x(x^2-3)(x^2-1))/(x^2-1)=x^3-3x#

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

graph{(x^5-4x^3+3x)/(x^2-1) [-10, 10, -5, 5]}