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