在不使用计算器的情况下，解不等式
$\frac{x}{3x-4} > \frac{1}{x}.$

1. Rearrange the inequality to compare with zero and express it as a single algebraic fraction.

$\frac{x}{3x-4} - \frac{1}{x} > 0$\n$\frac{x^2 - (3x - 4)}{x(3x - 4)} > 0$\n$\frac{x^2 - 3x + 4}{x(3x - 4)} > 0$

2. Analyse the sign of the quadratic numerator.

For the numerator $x^2 - 3x + 4$:\nDiscriminant $\Delta = (-3)^2 - 4(1)(4) = 9 - 16 = -7$\nSince $\Delta < 0$ and the coefficient of $x^2$ is positive ($1 > 0$), we have $x^2 - 3x + 4 > 0$ for all real $x$.

3. Determine the solution by evaluating the sign of the denominator.

Since the numerator is always strictly positive, the entire fraction is positive if and only if the denominator is also positive:\n$x(3x - 4) > 0$\nThe critical values are $x = 0$ and $x = \frac{4}{3}$.\nSolving this quadratic inequality yields:\n$x < 0 \text{ or } x > \frac{4}{3}$