If the ratio of 7^th term from the beginning to the 7^th term from the end in the expansion of ${\left(\sqrt[3]{2}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt[3]{3}}}\right)}^x$  is $\frac{1}{6}$ then x is 

${\left(\sqrt[3]{2}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt[3]{3}}}\right)}^x$

7^th term from end $={(n-7+2)}^{\text{th}}\text{term from begining }$

$={(n-5)}^{\text{th}}\text{term from begining}$

$\frac{T_{6+1}}{T_{n-6+1}}=\frac{1}{6}$

$\Rightarrow \frac{{}^x{\text{C}}_6{\left(\sqrt[3]{2}\right)}^{x-6}.{\left(\frac{1}{\sqrt[3]{3}}\right)}^6}{{}^x{\text{C}}_{x-6}{\left(\sqrt[3]{2}\right)}^6.{\left(\frac{1}{\sqrt[3]{3}}\right)}^{x-6}}=\frac{1}{6}$

$\Rightarrow \frac{{}^x{\text{C}}_6}{{}^x{\text{C}}_{x-6}}.{\left(2^{\frac{{\displaystyle 1}}{{\displaystyle 3}}}\right)}^{x-12}.{\left(\frac{{\displaystyle 1}}{{\displaystyle 3^{\frac{{\displaystyle 1}}{{\displaystyle 3}}}}}\right)}^{12-x}=\frac{1}{6}$

$\Rightarrow 2^{\frac{x-12}{3}}{.3}^{\frac{x-12}{3}}=\frac{1}{6}$

$\Rightarrow 6^{\frac{x-12}{3}}=6^{-1}$

$\begin{array}{l}\Rightarrow x-12=-3\\ \Rightarrow x=9\end{array}$