y=3x-1/2x

My attempt:

$\displaystyle y=(6x^2-1)/(2x)$

x= $\displaystyle (6y^2-1)/(2y)$

$\displaystyle 2yx=6y^2-1$

$\displaystyle 2yx-6y^2=-1$

$\displaystyle

2y(x-3y)=-1$

$\displaystyle 2y=-1/(x-3y)$

( I get stuck her what do l do with the y in the denominator ? )