Answer
$$\frac{{dy}}{{dx}} = \frac{2}{{x\ln 10{{\left( {1 - \log x} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{1 + \log x}}{{1 - \log x}} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{1 + \log x}}{{1 - \log x}}} \right] \cr
& {\text{By the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 - \log x} \right)\left( {\frac{1}{{x\ln 10}}} \right) - \left( {1 + \log x} \right)\left( { - \frac{1}{{x\ln 10}}} \right)}}{{{{\left( {1 - \log x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{1}{{x\ln 10}} - \frac{{\log x}}{{x\ln 10}} + \frac{1}{{x\ln 10}} + \frac{{\log x}}{{x\ln 10}}}}{{{{\left( {1 - \log x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{2}{{x\ln 10}}}}{{{{\left( {1 - \log x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{2}{{x\ln 10{{\left( {1 - \log x} \right)}^2}}} \cr} $$
