The greatest value of the function ${\log }_{x}1/9-{\log }_3{x}^2$

$(x > 1)$is 

$\begin{array}{r}f\left(x\right)={\log }_x(1/9)-{\log }_3{x}^2=-\frac{\log 9}{\log x}-\frac{2\log x}{\log 3}\end{array}$

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=\frac{2\left((\log 3{)}^2-(\log x{)}^2\right)}{x(\log x{)}^2\log 3}\\ \end{array}$So $f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow \log x$

$\begin{array}{r}=\pm \log 3\\ \end{array}$

$\Rightarrow x=3$ or $1/3.$One can see by first derivative test

that max $f\left(x\right)=f\left(3\right)=-4$.