Answer
$$y' = \frac{{x{e^x}\ln x - {e^x}}}{{x{{\left( {\ln x} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{{e^x}}}{{\ln x}} \cr
& {\text{use quotient rule }} \cr
& \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr
& y' = \frac{{\left( {\ln x} \right)\left[ {{e^x}} \right]' - {e^x}\left[ {\ln x} \right]'}}{{{{\left( {\ln x} \right)}^2}}} \cr
& {\text{compute derivatives}} \cr
& y' = \frac{{\left( {\ln x} \right)\left( {{e^x}} \right) - {e^x}\left( {\frac{1}{x}} \right)}}{{{{\left( {\ln x} \right)}^2}}} \cr
& {\text{simplify}} \cr
& y' = \frac{{{e^x}\ln x - \frac{{{e^x}}}{x}}}{{{{\left( {\ln x} \right)}^2}}} \cr
& y' = \frac{{x{e^x}\ln x - {e^x}}}{{x{{\left( {\ln x} \right)}^2}}} \cr} $$
