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.

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