We note first that division by zero is not allowed and that for this reason \(6x+4\) may not be equal to zero and that therefore \(x=-{{2}\over{3}}\) is not a solution.

We now distinguish two cases, namely \(6x+4>0\) and \(6x+4<0\).
In both cases we multiply the inequality on both sides by \(6x+4\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(6x+4>0\), i.e. \(x> -{{2}\over{3}}\). Then we get \(3<-6(6x+4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(36x<-27\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{3}\over{4}}\).
So we have the following system of inequalities: \(x> -{{2}\over{3}}\,\wedge\; x < -{{3}\over{4}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(6x+4<0\), i.e. \(x< -{{2}\over{3}}\). Then we get \(3>-6(6x+4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(36x>-27\).
Then, division by the coefficient of \(x\) gives \(x > -{{3}\over{4}}\).
So we have the following system of inequalities: \(x< -{{2}\over{3}}\,\wedge\; x > -{{3}\over{4}}\)
and this simplifies to \(-{{3}\over{4}}\lt x\lt -{{2}\over{3}}\).

The solution of the original inequality is \(-{{3}\over{4}}\lt x\lt -{{2}\over{3}}\).