Step 1: Rewrite the given differential equation.

$$\left[x \sin^2\left(\frac{y}{x}\right) - y\right] dx + x dy = 0$$ $$x\, dy = \left[y - x \sin^2\left(\frac{y}{x}\right)\right] dx$$ $$\frac{dy}{dx} = \frac{y}{x} - \sin^2\left(\frac{y}{x}\right)$$

Step 2: Identify that this is a homogeneous differential equation since the RHS is a function of $\left(\frac{y}{x}\right)$.

Substitute $y = vx$, so:

$$\frac{dy}{dx} = v + x\frac{dv}{dx}$$

Step 3: Substitute and simplify.

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

Step 4: Integrate both sides.

$$\int \cosec^2(v)\, dv = -\int \frac{dx}{x}$$ $$-\cot(v) = -\ln|x| + C$$ $$\cot(v) = \ln|x| + C$$

Step 5: Replace $v = \frac{y}{x}$ to get the general solution.

$$\mathbf{\cot\left(\frac{y}{x}\right) = \ln|x| + C}$$

Step 6: Apply the particular condition $y = \frac{\pi}{4}$ when $x = 1$.

$$\cot\left(\frac{\pi}{4}\right) = \ln|1| + C$$ $$1 = 0 + C$$ $$\mathbf{C = 1}$$

Step 7: Write the particular solution.

$$\mathbf{\cot\left(\frac{y}{x}\right) = \ln|x| + 1}$$