Logging around!

Put $a=3^{\frac{1}{x}}$ and $b=2^{\frac{1}{x}}$ , in the given equation we get,

$6a^2-13ab+6b^2=0$ or, $(2a-3b)(3a-2b)=0$ .

So, $\frac{a}{b}=\frac{3}{2}^\frac{1}{x}=\frac{3}{2}\: or\: \frac{2}{3}$ .

Thus, $x=1,-1$ and product of the roots is ${-1}$ .

$6 \cdot (3^2)^{\frac{1}{x}} - 13 \cdot (2 \cdot 3)^{\frac{1}{x}} + 6 \cdot (2^2)^{\frac{1}{x}} = 0$

Consideremos $3^{\frac{1}{x}} = m$ e $2^{\frac{1}{x}}= n$ .

Segue,

$6m^2 - 13nm + 6n^2 = 0 \\\\ 6m^2 - 9mn - 4mn + 6n^2 = 0 \\\\ 3m(2m - 3n) - 2n(2m - 3n) = 0 \\\\ (3m - 2n)(2m - 3n) = 0$

Fator I,

$3m - 2n = 0 \\\\ 3m = 2n \\\\ 3 \cdot 3^{\frac{1}{x}} = 2 \cdot 2^{\frac{1}{x}} \\\\ (\frac{3}{2})^{\frac{1}{x}} = \frac{2}{3} \\\\ (\frac{3}{2})^{\frac{1}{x}} = (\frac{3}{2})^{- 1} \\\\ \frac{1}{x} = - 1 \\\\ \boxed{x = - 1}$

Fator II,

$2m - 3n = 0 \\\\ 2m = 3n \\\\ 2 \cdot 3^{\frac{1}{x}} = 3 \cdot 2^{\frac{1}{x}} \\\\ (\frac{3}{2})^{\frac{1}{x}} = \frac{3}{2} \\\\ \frac{1}{x} = 1 \\\\ \boxed{x = 1}$

Com efeito,

$\boxed{\boxed{x_1 \cdot x_2 = - 1}}$