You are right about separating it in three cases (the last one should say x < 0). For each of those cases you need to solve the inequality. In each case, you'll get a set of solutions. Then you need to take the intersection with the corresponding set of x's: take only positive solutions in the second case and only negative solutions in the third case. Written in terms of set operations, the final set of solutions is where A and B are sets of solutions in the second and third case, respectively. Equivalently, x is a solution to the original equation iff (1) x = 0, or (2) x > 0 and x is a solution in the second case, or (3) x < 0 and x is a solution in the third case. Note the distribution of "and"s and "or"s.

In practice, I would write the three cases separately and clearly note the side condition (x = 0, x > 0, x < 0) for each case. After solving each case, it is important not to forget and combine the solution with the side conditions using "and," and then combine the solutions for different cases using "or."

The last formula is ambiguous: does it mean or ? And both of these interpretations are wrong. In fact, who said that it is allowed to convert into ? I would just memorize that

is equivalent to

In this case, it can be verified in two ways. First,

Alternatively, you can move a point along the x-axis from 0 to the right or to the left and see where the inequality becomes true.

Remember that you need to take only the positive solutions from the second case.

No, it is not wrong. You need to take the negative x satisfying .