Answer
$$y = {\left( {\frac{3}{2}\sqrt {2x} + C} \right)^{2/3}}$$
Work Step by Step
$$\eqalign{
& \sqrt {2xy} \frac{{dy}}{{dx}} = 1 \cr
& {\text{Use the radical property }}\sqrt {ab} = \sqrt a \sqrt b \cr
& \sqrt {2x} \sqrt y \frac{{dy}}{{dx}} = 1 \cr
& {\text{Separate the variables}} \cr
& \sqrt y dy = \frac{1}{{\sqrt {2x} }}dx \cr
& {y^{1/2}}dy = \frac{1}{{\sqrt 2 {x^{1/2}}}}dx \cr
& {y^{1/2}}dy = \frac{1}{{\sqrt 2 }}{x^{ - 1/2}}dx \cr
& {\text{Integrate both sides of the equation}} \cr
& \frac{{{y^{3/2}}}}{{3/2}} = \frac{1}{{\sqrt 2 }}\left( {\frac{{{x^{1/2}}}}{{1/2}}} \right) + C \cr
& {\text{Simplify}} \cr
& \frac{2}{3}{y^{3/2}} = \sqrt 2 \sqrt x + C \cr
& \frac{2}{3}{y^{3/2}} = \sqrt {2x} + C \cr
& {\text{Solve for }}y \cr
& {y^{3/2}} = \frac{3}{2}\sqrt {2x} + C \cr
& y = {\left( {\frac{3}{2}\sqrt {2x} + C} \right)^{2/3}} \cr} $$
