Let [tex]y=x^{-1}[/tex], so that [tex]y^2=x^{-2}[/tex]. Then
[tex]6x^{-2}-7x^{-1}+1=0\iff6y^2-7y+1=0\implies (6y-1)(y-1)=0[/tex]
which has two solutions, [tex]y=\dfrac16[/tex] and [tex]y=1[/tex]. So the solutions in terms of [tex]x[/tex] would be
[tex]x^{-1}=\dfrac16\implies x=6[/tex]
[tex]x^{-1}=1\implies x=1[/tex]
