Answer
$$\frac{{dz}}{{dy}} = \frac{{{{10}^y}}}{{\left( {\ln 10} \right)y}}\left( {1 + y{{\left( {\ln 10} \right)}^2}\log y} \right)$$
Work Step by Step
$$\eqalign{
& z = {10^y}\log y \cr
& {\text{differentiate with respect to }}y \cr
& \frac{{dz}}{{dy}} = \frac{d}{{dy}}\left[ {{{10}^y}\log y} \right] \cr
& {\text{use product rule}} \cr
& \frac{{dz}}{{dy}} = {10^y}\frac{d}{{dy}}\left[ {\log y} \right] + \log y\frac{d}{{dy}}\left[ {{{10}^y}} \right] \cr
& {\text{compute derivatives using the basic rules in this section}} \cr
& \frac{{dz}}{{dy}} = {10^y}\left( {\frac{1}{{\left( {\ln 10} \right)y}}} \right) + \log y\left( {{{10}^y}\ln 10} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dz}}{{dy}} = \frac{{{{10}^y}}}{{\left( {\ln 10} \right)y}} + {10^y}\left( {\ln 10} \right)\log y \cr
& \frac{{dz}}{{dy}} = {10^y}\left( {\frac{1}{{\left( {\ln 10} \right)y}} + \left( {\ln 10} \right)\log y} \right) \cr
& \frac{{dz}}{{dy}} = \frac{{{{10}^y}}}{{\left( {\ln 10} \right)y}}\left( {1 + y{{\left( {\ln 10} \right)}^2}\log y} \right) \cr} $$
