This equation is separable, as
[tex]\dfrac{\mathrm dy}{\mathrm dx}=y(y-2)e^x\implies\dfrac{\mathrm dy}{y(y-2)}=e^x\,\mathrm dx[/tex]
Integrate both sides; on the left, expand the fraction as
[tex]\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)[/tex]
Then
[tex]\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C[/tex]
[tex]\implies\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x+C[/tex]
Since [tex]y(0)=1[/tex], we get
[tex]\dfrac12\ln\left|\dfrac{1-2}1\right|=e^0+C\implies C=-1[/tex]
so that the particular solution is
[tex]\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x-1\implies\boxed{y=\dfrac2{1-e^{2e^x-2}}}[/tex]
