Answer
$$y' = \frac{{2x\ln 10 + 2x\ln 10\log x - x}}{{{{\left( {1 + \log x} \right)}^2}\ln 10}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^2}}}{{1 + \log x}} \cr
& {\text{quotient rule }} \cr
& \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr
& y' = \frac{{\left( {1 + \log x} \right)\left( {{x^2}} \right)' - {x^2}\left( {1 + \log x} \right)'}}{{{{\left( {1 + \log x} \right)}^2}}} \cr
& y' = \frac{{\left( {1 + \log x} \right)\left( {2x} \right) - {x^2}\left( {\frac{1}{{x\ln 10}}} \right)}}{{{{\left( {1 + \log x} \right)}^2}}} \cr
& {\text{simplify}} \cr
& y' = \frac{{2x + 2x\log x - \frac{x}{{\ln 10}}}}{{{{\left( {1 + \log x} \right)}^2}}} \cr
& y' = \frac{{2x\ln 10 + 2x\ln 10\log x - x}}{{{{\left( {1 + \log x} \right)}^2}\ln 10}} \cr} $$
