1. Rearrange the inequality to have zero on one side and combine into a single algebraic fraction.

$\frac{3x^2 + 1}{2x^2 + 3x - 2} - 1 > 0$
$\frac{3x^2 + 1 - (2x^2 + 3x - 2)}{2x^2 + 3x - 2} > 0$
$\frac{x^2 - 3x + 3}{2x^2 + 3x - 2} > 0$

2. Analyse the sign of the numerator.

For the numerator $x^2 - 3x + 3$:
Discriminant $\Delta = (-3)^2 - 4(1)(3) = 9 - 12 = -3$.
Since $\Delta < 0$ and the coefficient of $x^2$ is positive ($1 > 0$), $x^2 - 3x + 3 > 0$ for all real values of $x$.

(Alternatively, by completing the square: $x^2 - 3x + 3 = (x - \frac{3}{2})^2 + \frac{3}{4} > 0$ for all $x$.)

3. Deduce the condition for the denominator and solve the resulting quadratic inequality.

Since the numerator is strictly positive, the denominator must also be positive for the entire fraction to be greater than zero:
$2x^2 + 3x - 2 > 0$
Factorising the quadratic expression:
$(2x - 1)(x + 2) > 0$
Considering the roots $x = \frac{1}{2}$ and $x = -2$, the parabola opens upwards. Thus, the expression is positive outside the roots:
$x < -2 \text{ or } x > \frac{1}{2}$

1. Solve the inequality in part (i) to establish the necessary base result.

$\frac{3x^2 + 1}{2x^2 + 3x - 2} - 1 > 0$
$\frac{x^2 - 3x + 3}{(2x - 1)(x + 2)} > 0$
$\text{Since } x^2 - 3x + 3 = \left(x - \frac{3}{2}\right)^2 + \frac{3}{4} > 0 \text{ for all real } x,$
$(2x - 1)(x + 2) > 0$
$x < -2 \quad \text{or} \quad x > \frac{1}{2}$

2. Relate the target inequality to part (i) by applying a modulus substitution.

Notice that $x^2 = |x|^2$, so the inequality can be rewritten as:
$\frac{3|x|^2 + 1}{2|x|^2 + 3|x| - 2} > 1$
This is identical to the inequality in (i) with $x$ replaced by $|x|$. Using the established result:
$|x| < -2 \quad \text{or} \quad |x| > \frac{1}{2}$

3. Solve the resulting modulus inequalities, rejecting any invalid cases.

Since $|x| \ge 0$ for all real $x$, the inequality $|x| < -2$ has no real solutions.
We only solve $|x| > \frac{1}{2}$:
$x < -\frac{1}{2} \quad \text{or} \quad x > \frac{1}{2}$