I would surely like to know your approach

Let $2^x$ be $a$ and $3^x$ be $b$ .Therefore $8^x=a^3$ and $27^x=b^3$

$12^x=(2^2×3)^{x}=(2^{x})^{2}.3^{x}=a^{2}b~~ \text {similarly }18^{x}=ab^{2}$ The equation becomes $\dfrac {a^3+b^3}{ab (a+b)}=\dfrac{7}{6}$ As $a^3+b^3=(a+b)(a^2+b^2-ab)$ $6 (a^2+b^2-ab)=7ab \\ \implies 6a^2-13ab+6b^2=0$ Divide throughout by $b^2$ $\implies 6 \left (\dfrac {a}{b}\right)^{2}-13\left (\dfrac {a}{b}\right)+6=0$ The equation becomes a quadratic in $\left (\dfrac {a}{b}\right)$ .Solving for $\left (\dfrac {a}{b}\right)$ ,we get $\left (\dfrac{a}{b}\right)=\dfrac{2}{3},\dfrac {3}{2}\\ \implies \left (\dfrac {2}{3}\right)^ {x}=\dfrac {2}{3} \text {or} \dfrac {3}{2} \\ \boxed {\therefore x=\pm 1}$

Canceling out $2^x+3^x$ , we can write the given equation as $\frac{4^x-6^x+9^x}{6^x}=\left(\frac{2}{3}\right)^x-1+\left(\frac{3}{2}\right)^x=\frac{7}{6}$ and, after multiplying with $\left(\frac{2}{3}\right)^x$ , $\left(\left(\frac{2}{3}\right)^x-\frac{2}{3}\right)\left(\left(\frac{2}{3}\right)^x-\frac{3}{2}\right)=0$ The solutions are $x=1,-1$ , with their product being $\boxed{-1}$ .