Answer
$${\text{Separable}}$$
Work Step by Step
$$\eqalign{
& \sqrt x \frac{{dy}}{{dx}} = \frac{{1 + \ln x}}{y} \cr
& {\text{divide both sides by }}\sqrt x \cr
& \frac{{dy}}{{dx}} = \frac{{1 + \ln x}}{{y\sqrt x }} \cr
& \frac{{dy}}{{dx}} = \frac{{1 + \ln x}}{{y{x^{1/2}}}} \cr
& {\text{or}} \cr
& \frac{{dy}}{{dx}} = \frac{{{x^{ - 1/2}}\left( {1 + \ln x} \right)}}{y} \cr
& {\text{The equation can be written in the form }}\frac{{dy}}{{dx}} = \frac{{p\left( x \right)}}{{q\left( y \right)}},{\text{ then}} \cr
& {\text{the given equation is separable}} \cr
& \cr
& \sqrt x \frac{{dy}}{{dx}} = \frac{{1 + \ln x}}{y} \cr
& \sqrt x \frac{{dy}}{{dx}} = \frac{1}{y} + \frac{{\ln x}}{y} \cr
& \sqrt x \frac{{dy}}{{dx}} - \frac{1}{y} = \frac{{\ln x}}{y} \cr
& {\text{divide by }}\sqrt x \cr
& \frac{{dy}}{{dx}} - \frac{1}{{y\sqrt x }} = \frac{{\ln x}}{{y\sqrt x }} \cr
& {\text{The equation cannot be written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\,\,\,\left( {{\text{linear form}}} \right), \cr
& {\text{ then}}{\text{, the given equation is not linear}} \cr} $$