More function problems...

Find the points on the curve y = 2/3x + 2 where the tangent is parallel to the line y = (-3/2x) - 1

I know the slope of the tangent is -3/2 as it is parallel to the line y = (-3/2x) - 1.

Therefore m = -3/2

So I can express the curve as...

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

Using a combination of the product rule and chain rule I have come up with:

= -6/(3x - 2)^2 + 1/(3x - 2)

So...

-3/2 = -6/(3x - 2)^2 + 1/(3x - 2)

But that's where I get hung up. Can I express that as the following then solve for x?

-3/2 = -5/(3x - 2)^2 + (3x - 2)

Thanks!