The solution is: $xy \cos(y/x) = C$

Step 1: Rewriting the differential equation

The given equation is:

$$\{x \cos(y/x) + y \sin(y/x)\} y \, dx = \{y \sin(y/x) - x \cos(y/x)\} x \, dy$$

Rearranging to get $\frac{dy}{dx}$ form:

$$\frac{dy}{dx} = \frac{\{x \cos(y/x) + y \sin(y/x)\} y}{\{y \sin(y/x) - x \cos(y/x)\} x}$$

Step 2: Showing the equation is homogeneous

Let $y = vx$, so $y/x = v$.

Substituting in the equation:

$$\frac{dy}{dx} = \frac{[x \cos(v) + vx \sin(v)] \, vx}{[vx \sin(v) - x \cos(v)] \, x}$$ $$= \frac{x^2 v \{\cos(v) + v \sin(v)\}}{x^2 \{v \sin(v) - \cos(v)\}}$$ $$= \frac{v \{\cos(v) + v \sin(v)\}}{v \sin(v) - \cos(v)}$$

Since the right side depends only on $v = y/x$, the equation is homogeneous.

Step 3: Solving using substitution

Let $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$

$$v + x\frac{dv}{dx} = \frac{v\{\cos(v) + v \sin(v)\}}{v \sin(v) - \cos(v)}$$ $$x\frac{dv}{dx} = \frac{v\{\cos(v) + v \sin(v)\} - v\{v \sin(v) - \cos(v)\}}{v \sin(v) - \cos(v)}$$ $$x\frac{dv}{dx} = \frac{v \cos(v) + v^2 \sin(v) - v^2 \sin(v) + v \cos(v)}{v \sin(v) - \cos(v)}$$ $$x\frac{dv}{dx} = \frac{2v \cos(v)}{v \sin(v) - \cos(v)}$$

Step 4: Separating variables

$$\frac{v \sin(v) - \cos(v)}{2v \cos(v)} \, dv = \frac{dx}{x}$$ $$\frac{1}{2}\left[\tan(v) - \frac{1}{v}\right] dv = \frac{dx}{x}$$

Step 5: Integrating both sides

$$\frac{1}{2}\left[\ln|\sec(v)| - \ln|v|\right] = \ln|x| + C$$ $$\ln\left|\frac{\sec(v)}{v}\right| = 2\ln|x| + C'$$ $$\frac{\sec(v)}{v} = C'' x^2$$ $$\frac{1}{v \cos(v)} = C'' x^2$$ $$v \cos(v) = \frac{C'''}{x^2}$$

Step 6: Substituting back $v = y/x$

$$\frac{y}{x} \cos\left(\frac{y}{x}\right) = \frac{C'''}{x^2}$$ $$y \cos\left(\frac{y}{x}\right) = \frac{C'''}{x}$$

∴ $xy \cos(y/x) = C$ (where $C$ is an arbitrary constant)