Answer
$$\frac{{dy}}{{dr}} = \frac{1}{r}\left( {\frac{{{{\log }_3}r}}{{\ln 9}} + \frac{{{{\log }_9}r}}{{\ln 3}}} \right)$$
Work Step by Step
$$\eqalign{
& y = {\log _3}r \cdot {\log _9}r \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}r \cr
& \frac{{dy}}{{dr}} = \frac{d}{{dx}}\left[ {{{\log }_3}r \cdot {{\log }_9}r} \right] \cr
& {\text{Use the product rule}} \cr
& \frac{{dy}}{{dr}} = {\log _3}r\frac{d}{{dr}}\left[ {{{\log }_9}r} \right] + {\log _9}r\frac{d}{{dr}}\left[ {{{\log }_3}r} \right] \cr
& {\text{Use the rule }}\frac{d}{{dx}}\left[ {{{\log }_a}x} \right] = \frac{1}{{\left( {\ln a} \right)x}}\cr
&{\text{Then}} \cr
& \frac{{dy}}{{dr}} = {\log _3}r\left( {\frac{1}{{\left( {\ln 9} \right)r}}} \right) + {\log _9}r\left( {\frac{1}{{\left( {\ln 3} \right)r}}} \right) \cr
& {\text{Simplifying, we get:}} \cr
& \frac{{dy}}{{dr}} = \frac{{{{\log }_3}r}}{{\left( {\ln 9} \right)r}} + \frac{{{{\log }_9}r}}{{\left( {\ln 3} \right)r}} \cr
& \frac{{dy}}{{dr}} = \frac{1}{r}\left( {\frac{{{{\log }_3}r}}{{\ln 9}} + \frac{{{{\log }_9}r}}{{\ln 3}}} \right) \cr} $$
