$\text{ The solution of the equation }\left(x^2+y^2\right)\mathrm{d}y=xy \mathrm{d}x\text{ is }y=y\left(x\right).\text{ If }y\left(1\right)=1\text{ and }y\left(x_0\right)=e\text{ , }$$\text{ then }{x}_0\text{ is }$

$\text{ Given differential equation is }\left(x^2+y^2\right)\mathrm{d}y=xy \mathrm{d}x$

$\Rightarrow \frac{\text{d}y}{\text{d}x}=\frac{xy}{x^2+y^2}\mathrm{.......}\left(1\right)$

It is a homogenous equation 

$\text{ Put }y=vx$

$\Rightarrow \frac{\text{d}y}{\text{d}x}=v+x\frac{\text{d}v}{\text{d}x}$

$\text{ From (1), }v+x\frac{\mathrm{d}v}{ \mathrm{d}x}=\frac{v}{1+v^2}$

$\Rightarrow \left(\frac{1+v^2}{v^3}\right)\text{\hspace{0.17em}d}v=\text{d}x$

$\Rightarrow \log v-\frac{1}{2v^2}=-\log x+c$

$\Rightarrow \log \left(\frac{y}{x}\right)-\frac{x^2}{2y^2}=-\log x+c$

$\Rightarrow \log y-\frac{x^2}{2y^2}=c\text{ and given }y\left(1\right)=1$

$\Rightarrow 0-\frac{1}{2}=c$

$\therefore \log y-\frac{x^2}{2y^2}=-\frac{1}{2}$

$\text{ Now }y\left(x_0\right)=e$

$\Rightarrow \log e-\frac{x_0^2}{2e^2}=-\frac{1}{2}$

$\Rightarrow x_0=\sqrt{3}e$