Answer
$${y^2} = \frac{1}{{2x + C{x^2}}}$$
Work Step by Step
$$\eqalign{
& xy' + y = x{y^3} \cr
& y' + \frac{1}{x}y = {y^3} \cr
& {\text{The differential equation is written in the form}} \cr
& y' + P\left( x \right)y = Q\left( x \right){y^n} \cr
& y' + \frac{1}{x}y = {y^3} \to P\left( x \right) = \frac{1}{x},{\text{ }}Q\left( x \right) = 1,{\text{ }}n = 3 \cr
& {\text{From the book we have that the general solution of a Bernoulli }} \cr
& {\text{equation is:}} \cr
& {y^{1 - n}}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} = \int {\left( {1 - n} \right)Q\left( x \right){e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }}dx} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Find }}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} \cr
& {e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} = {e^{\int {\left( {1 - 3} \right)\left( {\frac{1}{x}} \right)dx} }} = {e^{ - 2\int {\frac{1}{x}} dx}} = {e^{ - 2\ln \left| x \right|}} = \frac{1}{{{x^2}}} \cr
& {\text{Substituting }}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }},{\text{ }}n,{\text{ and }}Q\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {y^{1 - 3}}\left( {\frac{1}{{{x^2}}}} \right) = \int {\left( {1 - 3} \right)\left( 1 \right)\left( {\frac{1}{{{x^2}}}} \right)dx} + C \cr
& {y^{ - 2}}\left( {\frac{1}{{{x^2}}}} \right) = - 2\int {\frac{1}{{{x^2}}}dx} + C \cr
& {\text{Integrating}} \cr
& {y^{ - 2}}\left( {\frac{1}{{{x^2}}}} \right) = \frac{2}{x} + C \cr
& {\text{Solve for }}y \cr
& {y^{ - 2}} = \left( {\frac{2}{x}} \right){x^2} + C{x^2} \cr
& \frac{1}{{{y^2}}} = 2x + C{x^2} \cr
& {y^2} = \frac{1}{{2x + C{x^2}}} \cr} $$
