Show that the differential equation is a homogeneous equation and find its solution:
$\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y \, dx = \left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x \, dy$

The given differential equation is:
$\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y \, dx = \left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x \, dy$
$\frac{dy}{dx} = \frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x} \quad \dots(1)$
Let $F(x, y) = \frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x}$
$F(\lambda x, \lambda y) = \frac{\left\{\lambda x \cos \left(\frac{\lambda y}{\lambda x}\right)+\lambda y \sin \left(\frac{\lambda y}{\lambda x}\right)\right\} \lambda y}{\left\{\lambda y \sin \left(\frac{\lambda y}{\lambda x}\right)-\lambda x \cos \left(\frac{\lambda y}{\lambda x}\right)\right\} \lambda x} = \frac{\lambda^2 \left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\lambda^2 \left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x} = \lambda^0 F(x, y)$
Since $F(\lambda x, \lambda y) = \lambda^0 F(x, y)$,the equation is homogeneous.
Substitute $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
$v + x \frac{dv}{dx} = \frac{(x \cos v + vx \sin v) vx}{(vx \sin v - x \cos v) x} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v}$
$x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v} - v = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v} = \frac{2v \cos v}{v \sin v - \cos v}$
$\left(\frac{v \sin v - \cos v}{v \cos v}\right) dv = \frac{2}{x} dx \implies (\tan v - \frac{1}{v}) dv = \frac{2}{x} dx$
Integrating both sides: $\int (\tan v - \frac{1}{v}) dv = \int \frac{2}{x} dx$
$\log(\sec v) - \log v = 2 \log x + \log C \implies \log\left(\frac{\sec v}{v}\right) = \log(Cx^2)$
$\frac{\sec v}{v} = Cx^2 \implies \sec v = Cx^2 v$
Substituting $v = \frac{y}{x}$: $\sec \left(\frac{y}{x}\right) = Cx^2 \left(\frac{y}{x}\right) = Cxy$
$\cos \left(\frac{y}{x}\right) = \frac{1}{Cxy} \implies xy \cos \left(\frac{y}{x}\right) = k$,where $k = \frac{1}{C}$.