Given $\frac{d y}{d x}=\frac{x y}{x^2+y^2}$. Put $y=v x ; \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}$
$\begin{aligned} & \therefore v+x \frac{d v}{d x}=\frac{(x)(v x)}{x^2+v^2 x^2} \\ & \Rightarrow v+x \cdot \frac{d v}{d x}=\frac{v}{1+v^2} \Rightarrow x \frac{d v}{d x}=\frac{-v^3}{1+v^2} \\ & \Rightarrow \frac{\left(1+v^2\right)}{v^3} d v=-\frac{d x}{x} \Rightarrow\left(\frac{1}{v^3}+\frac{1}{v}\right) d v=-\frac{d x}{x}\end{aligned}$
Integrating both sides, $\int \frac{d v}{v^3}+\int \frac{d v}{v}=-\int \frac{d x}{x}$
$\begin{aligned} & \Rightarrow-\frac{1}{2 v^2}+\log v=-\log x-\log c \\ & \Rightarrow-\frac{x^2}{2 y^2}+\log y=-\log c \Rightarrow \log c y=\frac{x^2}{2 y^2} \\ & \Rightarrow c y=e^{x^2 / 2 y^2} \Rightarrow c^2 y^2=e^{x^2 / y^2} \\ & \therefore y^2 a=e^{x^2 / y^2}, \text { where } c^2=a .\end{aligned}$