How to deal with this inequation with 3 absolute values? 
Solve:$$|x-2|+|3x+2|-x-2|x+4|\le -x-4$$

I am having trouble even starting with this inequation. Do I find out first the zeroes of each absolute value and then I would get intervals for each absolute value?

Rewrite your equation as
$$|x-2| + |3x+2| + 4 = 2|x+4|
\iff |-x+2| + |3x+2| + 4 = |2x+8|$$
Recall in the triangle inequality among $n$ real numbers $u_1,u_2,\ldots, u_n$,
$$|u_1| + \cdots + |u_n| \ge |u_1 + \cdots + u_n|,$$
the equality is achieved when and only when all non-zero $u_k$ have the same sign. Since
$$(-x + 2) + (3x +2) + 4 = 2x+8\quad\verb/and/\quad 4 > 0$$
the equation you have is equivalent to
$$-x + 2 \ge 0\;\;\verb/and/\;\; 3x + 2 \ge 0 \quad\iff\quad x \in \left[ -\frac23, 2\right]$$
