Answer
$$\frac{{dy}}{{dx}} = \frac{{{{\cos }^2}y\left( {x{e^x} + 1} \right)}}{x}$$
Work Step by Step
$$\eqalign{
& \tan y = {e^x} + \ln x \cr
& {\text{find the derivative using implicit differentiation; }} \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\tan y} \right] = \frac{d}{{dx}}\left[ {{e^x} + \ln x} \right] \cr
& {\text{solving the derivatives}}{\text{, use }}\frac{d}{{dx}}\left[ {\tan u} \right] = {\sec ^2}u\frac{{du}}{{dx}}{\text{ and }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dx}} \cr
& {\sec ^2}y\frac{d}{{dx}}\left[ y \right] = {e^x} + \frac{1}{x} \cr
& {\sec ^2}y\frac{{dy}}{{dx}} = \frac{{x{e^x} + 1}}{x} \cr
& {\text{solving for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{x{e^x} + 1}}{{x{{\sec }^2}y}} \cr
& or \cr
& \frac{{dy}}{{dx}} = \frac{{{{\cos }^2}y\left( {x{e^x} + 1} \right)}}{x} \cr} $$
