DEs

$\begin{aligned} (x^2+y^2) \ dy & = xy \ dx & \small \color{#3D99F6} \text{Divide both sides by }xy \\ \left(\frac xy + \frac yx \right) \ dy & = dx \\ \left(\frac xy + \frac yx \right) \frac {dy}{dx} & = 1 & \small \color{#3D99F6} \text{Let } y = vx \implies \frac {dy}{dx} = v + x\frac {dv}{dx} \\ \left(\frac 1v + v \right) \left(v + x\frac {dv}{dx}\right) & = 1 \\ v + x\frac {dv}{dx} & = \frac v{1+v^2} \\ x\frac {dv}{dx} & = \frac v{1+v^2} - v \\ x\frac {dv}{dx} & = - \frac {v^3}{1+v^2} \\ - \frac {1+v^2}{v^3} dv & = \frac {dx}x \\ - \int \left(\frac 1{v^3} + \frac 1v\right) dv & = \int \frac 1x \ dx \\ \frac 1{2v^2} - \ln v & = \ln x + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ \frac 1{2(1^2)} - \ln 1 & = \ln 1 + C & \small \color{#3D99F6} \text{when } x = 1, \ y = 1, \implies v = 1 \\ \implies C & = \frac 12 \\ \implies \frac 1{2v^2} - \ln v & = \ln x + \frac 12 & \small \color{#3D99F6} y(x_o) = e \implies v = \frac e{x_o} \\ \frac {x_o^2}{2e^2} - \ln e + \ln 1 & = \ln 1 + \frac 12 \\ \frac {x_o^2}{2e^2} - 1 & = \frac 12 \\ \implies x_o & = \boxed{\sqrt 3 e} \end{aligned}$

$\begin{aligned} x^2 \ dy+y^2 \ dy & = xy \ dx \\ x\left( x \ dy - y \ dx \right) +y^2 \ dy & = 0 \\ x^3 \ d \left( \dfrac{y}{x} \right) +y^2 \ dy & = 0 & \color{#3D99F6}{\small{ \text{divide both sides by } y^3 \text{ and integrate }}} \\ -\dfrac{1}{2} \cdot \dfrac{x^2}{y^2} + \ln{y} + c = 0 \\ \text{for } x=1 \hspace{4mm} y=1 \hspace{2mm},\hspace{2mm} c = \dfrac{1}{2} \\ \text{for } y=e \hspace{4mm} x_{o}=\sqrt{3}e \hspace{4mm} \end{aligned}$