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