Write the expression as a single logarithm with coefficient 1. Assume all variables represent positive real numbers.
$$-\frac{2}{3}{\log }_{6}6m^2+\frac{1}{2}{\log }_{6}36m^2$$

Step 1 ：Given the expression $-\frac{2}{3}{\log }_{6}6m^2+\frac{1}{2}{\log }_{6}36m^2$, we need to write it as a single logarithm with coefficient 1. Assume all variables represent positive real numbers.

Step 2 ：The properties of logarithms state that $n{\log }_{a}b={\log }_a{b}^n$ and ${\log }_{a}b+{\log }_{a}c={\log }_a(b\ast c)$. Therefore, we can rewrite the expression as a single logarithm by first applying the power rule to each term, and then applying the product rule to combine the two terms into one.

Step 3 ：Applying the power rule, the expression becomes ${\log }_6(6m^2{)}^{-\frac{2}{3}}+{\log }_6(36m^2{)}^{\frac{1}{2}}$.

Step 4 ：Then, applying the product rule, the expression becomes ${\log }_6(6m^2{)}^{-\frac{2}{3}}(36m^2{)}^{\frac{1}{2}}$.

Step 5 ：$\boxed{{\displaystyle {\log }_6(6m^2{)}^{-\frac{2}{3}}(36m^2{)}^{\frac{1}{2}}}}$ is the expression written as a single logarithm with coefficient 1.