If $x^{3}dy+xydx=x^{2}dy+2ydx,y\left(2\right)=e$ then $y(-1)=$

$x^{3}dy+xydx=x^{2}dy+2ydx$

$\Rightarrow \left(x^3-x^2\right)dy=(2y-xy)dx$

$\Rightarrow \frac{dy}{y}=\frac{(2-x)}{x^2(x-1)}dx=\left(\frac{1}{x-1}-\frac{1}{x}-\frac{2}{x^2}\right)dx$

$\therefore \ln y=\ln \frac{x-1}{x}+\frac{2}{x}+c$

$y\left(2\right)=e\Rightarrow 1=-\ln 2+1+c\Rightarrow c=\ln 2$

$\ln y=\ln \frac{x-1}{x}+\frac{2}{x}+\ln 2$

$x=-1\Rightarrow \ln y=\ln 4-2=\ln \frac{4}{e^2}\Rightarrow y=\frac{4}{e^2}$