Answer
$$\frac{{dy}}{{dx}} = \frac{{2x{e^{2x}}\ln x\left( {{x^2} + 1} \right) + 2{x^2}{e^{2x}} - {e^{2x}}\left( {{x^2} + 1} \right)}}{{x{{\left( {\ln x} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{\left( {{x^2} + 1} \right){e^{2x}}}}{{\ln x}} \cr
& {\text{differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\frac{{\left( {{x^2} + 1} \right){e^{2x}}}}{{\ln x}}} \right) \cr
& {\text{use quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\ln x\frac{d}{{dx}}\left( {\left( {{x^2} + 1} \right){e^{2x}}} \right) - \left( {{x^2} + 1} \right){e^{2x}}\frac{d}{{dx}}\left( {\ln x} \right)}}{{{{\left( {\ln x} \right)}^2}}} \cr
& {\text{use product rule for }}\frac{d}{{dx}}\left( {\left( {{x^2} + 1} \right){e^{2x}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{\ln x\left[ {\left( {{x^2} + 1} \right)\frac{d}{{dx}}\left( {{e^{2x}}} \right) + {e^{2x}}\frac{d}{{dx}}\left( {\left( {{x^2} + 1} \right)} \right)} \right] - \left( {{x^2} + 1} \right){e^{2x}}\frac{d}{{dx}}\left( {\ln x} \right)}}{{{{\left( {\ln x} \right)}^2}}} \cr
& {\text{find derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\ln x\left[ {2{e^{2x}}\left( {{x^2} + 1} \right) + {e^{2x}}\left( {2x} \right)} \right] - \left( {{x^2} + 1} \right)\left( {{e^{2x}}} \right)\left( {1/x} \right)}}{{{{\left( {\ln x} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{2{e^{2x}}\ln x\left( {{x^2} + 1} \right) + 2x{e^{2x}} - \left( {{x^2} + 1} \right)\left( {{e^{2x}}} \right)\left( {1/x} \right)}}{{{{\left( {\ln x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{2x{e^{2x}}\ln x\left( {{x^2} + 1} \right) + 2{x^2}{e^{2x}} - {e^{2x}}\left( {{x^2} + 1} \right)}}{{x{{\left( {\ln x} \right)}^2}}} \cr} $$
