Answer
$$dy = \frac{1}{{2\sqrt x }}\left( {1 - \frac{1}{x}} \right)dx$$
Work Step by Step
$$\eqalign{
& y = \sqrt x + \frac{1}{{\sqrt x }} \cr
& y = {x^{1/2}} + {x^{ - 1/2}} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^{1/2}} + {x^{ - 1/2}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{x^{ - 1/2}} - \frac{1}{2}{x^{ - 3/2}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }} - \frac{1}{{2{x^{3/2}}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }} - \frac{1}{{2x\sqrt x }} \cr
& {\text{Factor}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }}\left( {1 - \frac{1}{x}} \right) \cr
& {\text{Write in differential form }}dy = f'\left( x \right)dx \cr
& dy = \frac{1}{{2\sqrt x }}\left( {1 - \frac{1}{x}} \right)dx \cr} $$
