Satisfyingest

The given expression can be factored as $(x + 3y)^{2} + (3y + 2z)^{2}$ . As both of the squared terms are positive we know by the AM-GM inequality that the minimum will be achieved when $x + 3y = 3y + 2z \Longrightarrow x = 2z$ . In this case the given expression reduces to $2(2z + 3y)^{2}$ .

Next, the given condition $xyz = \dfrac{2}{3}$ becomes $2yz^{2} = \dfrac{2}{3} \Longrightarrow 3y = \dfrac{1}{z^{2}}$ , in which case $2(2z + 3y)^{2} = 2\left(2z + \dfrac{1}{z^{2}}\right)^{2}$ .

Now by the AM-GM inequality $2z + \dfrac{1}{z^{2}} = z + z + \dfrac{1}{z^{2}} \ge 3\sqrt[3]{z \times z \times \dfrac{1}{z^{2}}} = 3$ , where equality occurs when $z = \dfrac{1}{z^{2}} \Longrightarrow z = 1$ .

The desired minimum is then $2\left(2z + \dfrac{1}{z^{2}}\right)^{2} = 2 \times 3^{2} = \boxed{18}$ , occurring when $(x,y,z) = \left(2, \dfrac{1}{3}, 1\right)$ .