Answer
$(-\infty,-\frac{2}{3})\cup(0,\frac{3}{2})$.
Work Step by Step
Step 1. Rewrite the inequality $6x-5-\frac{6}{x}\lt0 \Longrightarrow \frac{(2x-3)(3x+2)}{x}\lt0$, identify boundary points as $x=-\frac{2}{3},0,\frac{3}{2}$.
Step 2. Form intervals $(-\infty,-\frac{2}{3}),(-\frac{2}{3},0),(0,\frac{3}{2}),(\frac{3}{2},\infty)$.
Step 3. Choose test values for each interval $x=-1,-1/3,1,2$.
Step 4. Test the inequality to get results: $True,\ False,\ True,\ False$
Step 5. We have the solution $(-\infty,-\frac{2}{3})\cup(0,\frac{3}{2})$.
