Given differential equation is
$$
\begin{array}{l}
y \sin \left(\frac{x}{y}\right) d x=\left\{x \sin \left(\frac{x}{y}\right)-y\right\} d y \\
\Rightarrow \quad \frac{d x}{d y}=\frac{x \sin \left(\frac{x}{y}\right)-y}{y \sin \left(\frac{x}{y}\right)}=\frac{x}{y}-\frac{1}{\sin \left(\frac{x}{y}\right)}
\end{array}
$$
On putting $v=\frac{x}{y} \Rightarrow x=y$
$\Rightarrow \quad \frac{d x}{d y}=v \cdot 1+y \frac{d v}{d y}$ in Eq. $0 .$ we get
$v+y \frac{d v}{d y}=v-\frac{1}{\sin v}$
$\Rightarrow \quad y \frac{d v}{d y}=-\frac{1}{\sin v}$
$\Rightarrow \quad-\int \sin v d v=\int \frac{d y}{y}$
(on integrating)
$\Rightarrow \quad \cos v=\log y+C$
$\Rightarrow \quad \cos \left(\frac{x}{y}\right)=\log y+C$
Given at $x=\frac{\pi}{4}, y=1$ then from $\mathrm{Eq}$. (i) $\Rightarrow \quad \cos \left(\frac{\pi}{4}\right)=\log (1)+C$
$\Rightarrow \quad\left(C=\frac{1}{\sqrt{2}}\right)$
On putting the value of $C$ in Eq. (0), we get $\cos \left(\frac{x}{y}\right)=\log _{e} y+\frac{1}{\sqrt{2}}$
Which is the required solution. So option D is correct.