Answer
$${y^2} = 1 + \frac{{C{\text{ }}}}{{{e^{{x^2}}}}}$$
Work Step by Step
$$\eqalign{
& y' + xy = x{y^{ - 1}} \cr
& {\text{The differential equation is written in the form}} \cr
& y' + P\left( x \right)y = Q\left( x \right){y^n} \cr
& y' + xy = x{y^{ - 1}} \to P\left( x \right) = x,{\text{ }}Q\left( x \right) = x,{\text{ }}n = - 1 \cr
& {\text{From the book we have that the general solution of a Bernoulli }} \cr
& {\text{equation is:}} \cr
& {y^{1 - n}}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} = \int {\left( {1 - n} \right)Q\left( x \right){e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }}dx} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Find }}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} \cr
& {e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} = {e^{\int {\left( {1 + 1} \right)\left( x \right)dx} }} = {e^{\int {2x} dx}} = {e^{{x^2}}} \cr
& {\text{Substituting }}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }},{\text{ }}n,{\text{ and }}Q\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {y^{1 + 1}}{e^{{x^2}}} = \int {\left( {1 + 1} \right)\left( x \right){e^{{x^2}}}dx} + C{\text{ }} \cr
& {y^2}{e^{{x^2}}} = \int {2x{e^{{x^2}}}dx} + C{\text{ }} \cr
& {\text{Integrating}} \cr
& {y^2}{e^{{x^2}}} = {e^{{x^2}}} + C{\text{ }} \cr
& {y^2} = \frac{{{e^{{x^2}}}}}{{{e^{{x^2}}}}} + \frac{{C{\text{ }}}}{{{e^{{x^2}}}}} \cr
& {\text{Simplifying}} \cr
& {y^2} = 1 + \frac{{C{\text{ }}}}{{{e^{{x^2}}}}} \cr} $$
