Answer
$y=\left\{ \dfrac{1}{7},1 \right\}$
Work Step by Step
Let $z=y^{-1}$. Then the given equation, $
2y^{-2}-16y^{-1}+14=0
,$ is equivalent to
\begin{array}{l}\require{cancel}
2z^2-16z+14=0
\\\\
\dfrac{2z^2-16z+14}{2}=\dfrac{0}{2}
\\\\
z^2-8z+7=0
\\\\
(z-7)(z-1)=0
\\\\
z=\{ 1,7 \}
.\end{array}
Since $z=y^{-1}$, then if $z=1$, then,
\begin{array}{l}\require{cancel}
1=y^{-1}
\\\\
1=\dfrac{1}{y}
\\\\
\dfrac{1}{1}=y
\\\\
1=y
.\end{array}
Since $z=y^{-1}$, then if $z=7$, then,
\begin{array}{l}\require{cancel}
7=y^{-1}
\\\\
7=\dfrac{1}{y}
\\\\
\dfrac{1}{7}=y
.\end{array}
Hence, $
y=\left\{ \dfrac{1}{7},1 \right\}
.$
