How to solve y′+y=exy23?

There is one obvious solution ${\displaystyle y\left(x\right)=0}$.
${\displaystyle y^{\prime }+y=e^x{y}^{\frac{2}{3}}}$
${\displaystyle y^{\prime }{e}^x+y{e}^x=e^{2x}{y}^{\frac{2}{3}}}$
${\displaystyle {\left(y{e}^x\right)}^{\prime }=e^{2x}{y}^{\frac{2}{3}}}$
Substitute ${\displaystyle u\left(x\right)=y\left(x\right)e^x}$
${\displaystyle u^{\prime }=u^{\frac{2}{3}}{e}^{\frac{4x}{3}}}$
${\displaystyle u^{-\frac{2}{3}}{u}^{\prime }=e^{\frac{4x}{3}}}$
We have divided by ${\displaystyle u^{\frac{2}{3}}}$, so this is ok only for points such that ${\displaystyle u\left(x\right)\ne 0}$.
${\displaystyle \frac{1}{3}{u}^{\frac{-2}{3}}{u}^{\prime }=\frac{1}{3}{e}^{\frac{4x}{3}}}$
${\displaystyle {\left(u^{\frac{1}{3}}\right)}^{\prime }={\left(\frac{\frac{4x}{3}}{4}\right)}^{\prime }}$
${\displaystyle u^{\frac{1}{3}}=\frac{e^{\frac{4x}{3}}}{4}+C}$
${\displaystyle u={(C+\frac{e^{\frac{4x}{3}}}{4})}^3}$
${\displaystyle y={(C+\frac{e^{\frac{4x}{3}}}{4})}^3{e}^{-x}}$
To get ${\displaystyle y=0}$, we need ${\displaystyle C=-\frac{1}{4}}$
${\displaystyle y={\left(\frac{e^{4\frac{x}{3}}-1}{4}\right)}^3{e}^{-x}}$

The idea is that in differential equations of the form ${\displaystyle y^{\prime }+g\left(x\right)y=f\left(x\right)y^{\alpha }(\alpha \ne 1)}$, the Bernoulli equations, you can do the variable change ${\displaystyle y=u^{\beta }}$, and your equation becomes
${\displaystyle \beta u^{\prime }{u}^{\beta -1}+g\left(x\right)u^{\beta }=f\left(x\right)u^{\beta \alpha }}$
Multiplying through by ${\displaystyle \frac{1}{\beta }{u}^{1-\beta }}$, this is the same as
${\displaystyle u^{\prime }+\frac{1}{\beta }g\left(x\right)u=\frac{1}{\beta }f\left(x\right)u^{\beta \alpha +1-\beta }}$
So the idea is that if you choose ${\displaystyle \beta }$ so that ${\displaystyle \beta \alpha +1-\beta =0}$, in other words, ${\displaystyle \beta =\frac{1}{1-\alpha }}$, your equation becomes first-order linear and you can solve it using first-order linear methods. In this case, ${\displaystyle \alpha =\frac{2}{3},\text{so}\beta =3}$ and you make the variable change ${\displaystyle y=u^3}$