jmc

Working on the first equation, we have from the difference of squares factorization that ${\log }_6(x-y)+{\log }_6(x+y)={\log }_6(x^2-y^2)=2$, so $x^2-y^2=6^2=36$. Using the change of base formula, the second equation becomes $$\frac{\log (5x)}{\log y}=2⟹\log (5x)=2\log y=\log y^2.$$Substituting that $y^2=x^2-36$, it follows that $\log (x^2-36)=\log y^2=2\log y=\log 5x$. Since the logarithm is a one-to-one function, it follows that $x^2-36=5x$, so $x^2-5x-36=(x-9)(x+4)=0$. Thus, $x=9,-4$, but the second does not work. Thus, our answer is $x=\boxed{9}$.