x^3 - 2x^2 - 3x > 0

=> x(x^2 - 2x - 3) > 0

=> x(x - 3)(x + 1) > 0

If you equate each factor to zero, you get 3 values of x which are

-1, 0 and 3

Divide the real number line into three parts breaking at -1, 0 and 3 as

(- ∞, - 1), (- 1, 0) and (0, ∞)

Now examine the validity of the inequality in each of these three intervals.

For x ∈ (- ∞, - 1), x < 0, (x - 3) < 0 and (x + 1) < 0

=>  x(x - 3)(x + 1) < 0

=> x does not belong to (- ∞, - 1).

For x ∈ (- 1, 0), x < 0, (x - 3) < 0 and (x + 1) > 0

=> x(x - 3)(x + 1) > 0

=> x ∈ (- 1, 0).

For x ∈ (0, 3), x > 0, (x - 3) < 0 and (x + 1) > 0

=> x(x - 3)(x + 1) < 0

=> x does not belong to (0, 3).

Conclusion :

x ∈ (- 1, 0).