The given inequality is:

$\left| \frac{1}{x} - 3 \right| > 6$

To solve this, we will follow the steps for solving an absolute value inequality.

Step 1: Break the absolute value inequality into two cases.

The general rule for absolute value inequalities $|A| > B$ is:

$A > B \quad \text{or} \quad A < -B$

Here, $A = \frac{1}{x} - 3$ and $B = 6$.

So, the two cases are:

Case 1:

$\frac{1}{x} - 3 > 6$

Case 2:

$\frac{1}{x} - 3 < -6$

Step 2: Solve each case.

Case 1: $\frac{1}{x} - 3 > 6$

Add 3 to both sides:

$\frac{1}{x} > 9$

Now, take the reciprocal of both sides (note: taking the reciprocal reverses the inequality sign):

$x < \frac{1}{9}$

Case 2: $\frac{1}{x} - 3 < -6$

Add 3 to both sides:

$\frac{1}{x} < -3$

Now, take the reciprocal of both sides (again, reverse the inequality sign):

$x > -\frac{1}{3}$

Step 3: Combine the solutions.

The combined solution is:

$x < \frac{1}{9} \quad \text{or} \quad x > -\frac{1}{3}$

Step 4: Exclude zero.

Since $x = 0$ makes the expression undefined (division by zero), we exclude $x = 0$.

Thus, the solution in interval notation is:

$(-\infty, -\frac{1}{3}) \cup (0, \frac{1}{9})$

Would you like more details or explanations on any part of this?

Here are some related questions for practice:

1. How would the solution change if the inequality were $\left| \frac{1}{x} - 3 \right| \leq 6$?
2. What happens if you multiply or divide by a negative number in inequalities?
3. How do you handle inequalities when there are two absolute value expressions?
4. Can you graph the solution of $(-\infty, -\frac{1}{3}) \cup (0, \frac{1}{9})$?
5. How do you solve rational inequalities like $\frac{1}{x} + 4 < 5$?

Tip: When working with absolute values, always remember to split the equation or inequality into two cases and solve each one separately.