Step 1: Rewrite the equation in dy/dx form.
The equation $(x\sin\frac{y}{x})\,dy = (y\sin\frac{y}{x} - x)\,dx$ gives $\frac{dy}{dx} = \frac{y\sin\frac{y}{x} - x}{x\sin\frac{y}{x}}$. This is homogeneous since every expression depends only on the ratio $\frac{y}{x}$.
Step 2: Use the substitution v = y/x.
Let $v = \frac{y}{x}$, so $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting: $x\sin v\!\left(v + x\frac{dv}{dx}\right) = vx\sin v - x$.
Step 3: Simplify and isolate the derivative term.
Dividing by $x$: $\sin v\!\left(v + x\frac{dv}{dx}\right) = v\sin v - 1$. Expanding: $v\sin v + x\sin v\frac{dv}{dx} = v\sin v - 1$. Subtracting $v\sin v$: \[x\sin v\frac{dv}{dx} = -1\]
Step 4: Separate variables and integrate.
Separating: $\sin v\,dv = -\frac{dx}{x}$. Integrating both sides: \[\int \sin v\,dv = -\int \frac{dx}{x} \implies -\cos v = -\ln|x| + C_0\] Multiplying by $-1$: $\cos v = \ln|x| + C$ (where $C = -C_0$).
Step 5: Substitute back v = y/x.
Replacing $v = \frac{y}{x}$: \[\cos\frac{y}{x} = \log x + C\]
Step 6: Compare to identify A.
The given form is $\cos\frac{y}{x} = A\log x + C$. Comparing with our solution: $A = 1$. \[\boxed{A = 1}\]