You must look in the ORIGINAL equation to determine final solutions.
In this case, since you are subtracting 3, the rest of the left-hand side had better be positive or you cannot get zero (0). Further, since the absolute value is positive, this constrains the remaining piece, x, to be positive.
I note two things:
1) 3s are wild. You could simplfy your life by a FIRST step...
x|3x+2| - 1 = 0
2) You did it twice. What was the point of moving the 3 to the right-hand side, just to move it back to the left so you could factor the whole expression?
If I were to do this problem, I might proceed like this:
A) x|9x+6| - 3 = 0
That three (3) is everywhere.
B) x|3x+2| - 1 = 0
x > 0 (by the argument up front.)
If 3x+2 > 0 ==> x > -3/2, and since we already know x > 0, we're just about done.
C) x(3x+2) - 1 = 0
Distributive Property
D) 3x^2 + 2x - 1 = 0
It has one positive solution. (Why do I know that? It's a visual inspection.)
2^2 + 4(3)(1) = 4 + 12 = 16 <== Whoops. I should have been able to factor it.
E) (3x - 1)(x + 1) = 0
F) x = 1/3 or x = -1 (and x = -1 is already thrown out.)
To answer your specific question, -1/3 is not a solution to the equation because it does NOT satisfy the equation!
If x= -3/3= -1, then x|9x+ 6|+ 3= (-1)(9(-1)+ 6|+ 3= (-1)(-9+ 6)+ 3= (-1)(-3)+ 3= 3+ 3= 6, not 0.
Another way of looking at it is that so that removing the absolute value is like squaring both sides of the equation- and that can introduce new values of x that satisfy the new equation but not the original one.