We have, $x \cos \left(\frac{y}{x}\right)(y d x+x d y)$ $=y \sin \frac{y}{x}(x d y-y d x)$
$\Rightarrow \quad x y \sin \left(\frac{y}{x}\right) d y-y^2 \sin \left(\frac{y}{x}\right) d x$ $=x y \cos \left(\frac{y}{x}\right) d x+x^2 \cos \left(\frac{y}{x}\right) d y$
$\Rightarrow\left[x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)\right] d y$ $=\left[x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)\right] d x$
$\Rightarrow \frac{d y}{d x}=\frac{x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)}{x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)}$
$\Rightarrow \frac{d y}{d x}=\frac{\frac{y}{x} \cos \left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)^2 \sin \left(\frac{y}{x}\right)}{\frac{y}{x} \sin \left(\frac{y}{x}\right)-\cos \left(\frac{y}{x}\right)}$
Put, $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore \quad v+x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v}$
$\Rightarrow \quad x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v}$
$\Rightarrow \quad \frac{v \sin v-\cos v}{v \cos v} d v=2 \frac{d x}{x}$
$\Rightarrow \quad \int\left(\tan v-\frac{1}{v}\right) d v=2 \int \frac{d x}{x}$
$\Rightarrow \quad \log |\sec v|-\log |v|=2 \log |x|+\log \left|C_1\right|$
$\Rightarrow \quad \log \left|\frac{\sec v}{v}\right|-\log |x|^2=\log \left|C_1\right|$
$\Rightarrow \quad \log \left|\frac{\sec v}{v x^2}\right|=\log \left|C_1\right|$
$\Rightarrow \quad \frac{\sec v}{v x^2}=C_1 \Rightarrow \frac{\sec \left(\frac{y}{x}\right)}{\frac{y}{x}\left(x^2\right)}=C_1$
$\Rightarrow \quad \frac{\sec \left(\frac{y}{x}\right)}{x y}=C_1 \Rightarrow \sec \left(\frac{y}{x}\right)=C_1 x y$
$\Rightarrow \quad \frac{1}{\cos \left(\frac{y}{x}\right)}=\frac{C}{x y} \quad\left(\because\right.$ where $\left.\frac{1}{C_1}=C\right)$