jmc

Let $a=2^x$ and $b=3^x.$ Then the given equation becomes $$\frac{a^3+b^3}{a^{2}b+a{b}^2}=\frac{7}{6}.$$We can factor, to get $$\frac{(a+b)(a^2-ab+b^2)}{ab(a+b)}=\frac{7}{6}.$$Since $a$ and $b$ are positive, $a+b$ must be positive, so we can safely cancel the factors of $a+b,$ to get $$\frac{a^2-ab+b^2}{ab}=\frac{7}{6}.$$Then $6a^2-6ab+6b^2=7ab,$ which simplifies to $6a^2-13ab+6b^2=0.$ This equation factors as $(2a-3b)(3a-2b)=0,$ so $2a=3b$ or $3a=2b.$

If $2a=3b,$ then $2^{x+1}=3^{x+1},$ or $$\frac{2^{x+1}}{3^{x+1}}={\left(\frac{2}{3}\right)}^{x+1}=1.$$The only solution here is $x=-1.$

If $3a=2b,$ then $3\cdot 2^x=2\cdot 3^x,$ or $$\frac{3\cdot 2^x}{2\cdot 3^x}={\left(\frac{2}{3}\right)}^{x-1}=1.$$The only solution here is $x=1.$

Therefore, the solutions are $\boxed{-1,1}.$