Answer
$$\frac{{dy}}{{dx}} = \frac{{y\left( {x{e^{x + y}} - 1} \right)}}{{x\left( {1 - y{e^{x + y}}} \right)}}$$
Work Step by Step
$$\eqalign{
& \ln xy = {e^{x + y}} \cr
& {\text{use the logarithmic property: }}\ln ab = \ln a + \ln b \cr
& \ln x + \ln y = {e^{x + y}} \cr
& {\text{find the derivative using the implicit differentiation;}} \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\ln x} \right] + \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {{e^{x + y}}} \right] \cr
& {\text{solving derivatives}} \cr
& \frac{1}{x} + \frac{1}{y}\frac{{dy}}{{dx}} = {e^{x + y}}\frac{d}{{dx}}\left[ {x + y} \right] \cr
& \frac{1}{x} + \frac{1}{y}\frac{{dy}}{{dx}} = {e^{x + y}}\left( {1 + \frac{{dy}}{{dx}}} \right) \cr
& \frac{1}{x} + \frac{1}{y}\frac{{dy}}{{dx}} = {e^{x + y}} + {e^{x + y}}\frac{{dy}}{{dx}} \cr
& {\text{solving for }}\frac{{dy}}{{dx}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} - {e^{x + y}}\frac{{dy}}{{dx}} = {e^{x + y}} - \frac{1}{x} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} - {e^{x + y}}\frac{{dy}}{{dx}} = \frac{{x{e^{x + y}} - 1}}{x} \cr
& \left( {1 - y{e^{x + y}}} \right)\frac{{dy}}{{dx}} = \frac{{y\left( {x{e^{x + y}} - 1} \right)}}{x} \cr
& \frac{{dy}}{{dx}} = \frac{{y\left( {x{e^{x + y}} - 1} \right)}}{{x\left( {1 - y{e^{x + y}}} \right)}} \cr} $$
