Answer
$$dy = \frac{1}{{3\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}dx$$
Work Step by Step
$$\eqalign{
& y = \frac{{2\sqrt x }}{{3\left( {1 + \sqrt x } \right)}} \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{2\sqrt x }}{{3\left( {1 + \sqrt x } \right)}}} \right] \cr
& {\text{Use the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{3\left( {1 + \sqrt x } \right)\left( {2\left( {\frac{1}{{2\sqrt x }}} \right)} \right) - 2\sqrt x \left( 3 \right)\left( {\frac{1}{{2\sqrt x }}} \right)}}{{9{{\left( {1 + \sqrt x } \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 + \sqrt x } \right)\left( {\frac{1}{{\sqrt x }}} \right) - 1}}{{3{{\left( {1 + \sqrt x } \right)}^2}}} \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{{1 + \sqrt x - \sqrt x }}{{3\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{3\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}} \cr
& {\text{Write in differential form}} \cr
& dy = \frac{1}{{3\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}dx \cr} $$
