Absolutely

I let $y =\left| x \right|$ and solved the quadratic equation $0 = (y-1)(y-2)$ . That yields $y=1, y=2$ . Substitute back: $\left|x \right| = 1 \rightarrow x= \pm 1$ and $\left| x\right| = 2 \rightarrow x= \pm 2$ .

The given equation either has a solution or not.

In case it does not, then the sum of the values of the solution set is $0$ .

In case it does, let it be $x = a$

$\Rightarrow$ $\frac {2}{|a|} + |a| = 3$

We can claim that $x = -a$ is also a solution, because

$\frac {2}{|-a|} + |-a| = \frac {2}{|a|} + |a| = 3$

Hence for every solution $x = a$ , there exists a solution $x = -a$ which gives the sum of the values of the solution set to be $0$ , anyways.

There is no need to find the solutions, I think.

Multiplying with $|x|$ on both sides gives $2+|x|^2=3|x|$ , or simply $(|x|-1)(|x|-2) = 0$ . From this equation, we get $x = -2, -1, 1, 2$ , and the sum of possible values is zero.