Whoops, you used a different way of finding the inverse than I'm used to, so I got mixed up.

What I normally do is:

y = x/(3x - 2)
y(3x - 2) = x
3xy - 2y = x
3xy - x = 2y
x(3y - 1) = 2y
x = 2y / (3y - 1)

What you did was pretty much the same, but you switched the x and y variables, as it is standard to want to solve for y:

x = y/(3y - 2)
x(3y - 2) = y
3xy - 2x = y
3xy - y = 2x
y(3x - 1) = 2x
y = 2x / (3x-1)

As you can see, you come up with the same answer, just different names for the variables.  In mine, x is dependant on y, in yours (as it normally is), y is dependant on x.

x(3y - 2) = y
x = y / (3y - 2)