Answer
$2log_{6}x-log_{6}(x+3)$
Work Step by Step
We know that $log_{b}\frac{x}{y}=log_{b}x-log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $log_{6}\frac{x^{2}}{x+3}=log_{6}x^{2}-log_{6}(x+3)$.
We know that $log_{b}x^{r}=rlog_{b}x$ (where $x$ and $b$ are positive real numbers, $b\ne1$, and $r$ is a real number).
Therefore, $log_{6}x^{2}-log_{6}(x+3)=2log_{6}x-log_{6}(x+3)$.