Solve for x
x\in \left(\frac{1}{2},\frac{2}{3}\right)
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1-2x<0 3x-2<0
For the product to be positive, 1-2x and 3x-2 have to be both negative or both positive. Consider the case when 1-2x and 3x-2 are both negative.
x\in \left(\frac{1}{2},\frac{2}{3}\right)
The solution satisfying both inequalities is x\in \left(\frac{1}{2},\frac{2}{3}\right).
3x-2>0 1-2x>0
Consider the case when 1-2x and 3x-2 are both positive.
x\in \emptyset
This is false for any x.
x\in \left(\frac{1}{2},\frac{2}{3}\right)
The final solution is the union of the obtained solutions.
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