`f(x)=(x^(2)+x+1)/(x^(2)+x-1) {x^(2)+x+1` and `x^(2)+x-1` have no common factor`}`
`y=(x^(2)+x+1)/(x^(2)x-1)`
`implies yx^(2)+yx-y=x^(2)+x+1`
`implies(y-1)x^(2)+(y-1)x-y-1=0`
If `y=1`, the the above equation reduces to `-2=0`. Which is not true.
Further if `y!=1` then `(y-1)x^(2)+(y-1)x-y-1=0` is a quadratic and has real roots if
`(y-1)^(2)-r(y-1)(-y-1)ge0`
i.e. if `yle-3//5` or `yge1` but `y!=1`
Thus the range is `(-oo,-3//5]uu(1,oo)`
