Simplify the expression: ${\log }_{4}3\cdot {\log }_{243}64$

Using the base change formula ${\log }_{b}a=\frac{\ln a}{\ln b}$, rewrite the terms:
$${\log }_{4}3=\frac{\log 3}{\log 4}=\frac{\log 3}{\log 2^2}=\frac{\log 3}{2\log 2}$$

$${\log }_{243}64=\frac{\log 64}{\log 243}=\frac{\log 2^6}{\log 3^5}=\frac{6\log 2}{5\log 3}$$

Now, multiply the two simplified expressions:

$${\log }_{4}3\cdot {\log }_{243}64=\left(\frac{\log 3}{2\log 2}\right)\cdot \left(\frac{6\log 2}{5\log 3}\right)$$

Cancel the common terms $\log 3$ and $\log 2$:

$$\text{Result}=\frac{1}{2}\cdot \frac{6}{5}=\frac{6}{10}=\frac{3}{5}$$