Given differential equation is
$y \sin \left(\frac{x}{y}\right) d x=\left\{x \sin \left(\frac{x}{y}\right)-y\right\} d y$
$\Rightarrow \frac{d x}{d y}=\frac{x \sin \left(\frac{x}{y}\right)-y}{y \sin \left(\frac{x}{y}\right)}=\frac{x}{y}-\frac{1}{\sin \left(\frac{x}{y}\right)}$ ... (i)
On putting $v=\frac{x}{y} \Rightarrow x=v$
$ \Rightarrow \frac{d x}{d y} =v \cdot 1+y \frac{d v}{d y} $ in Eq. (i), we get
$ v+ y \frac{d v}{d y} =v-\frac{1}{\sin v} $
$\Rightarrow y \frac{d v}{d y} =-\frac{1}{\sin v} $
$\Rightarrow -\int \sin v d v=\int \frac{d y}{y}$ (on integrating)
$\Rightarrow \cos v=\log y+C $
$ \Rightarrow \cos \left(\frac{x}{y}\right)=\log y+C ... (i) $
Given at $x=\frac{\pi}{4}, y=1$ then from Eq. (i)
$\Rightarrow \cos \left(\frac{\pi}{4}\right)=\log (1)+C $
$\Rightarrow \left(C=\frac{1}{\sqrt{2}}\right)$
On putting the value of $C$ in Eq. (i), we get
$\cos \left(\frac{x}{y}\right)=\log _{e} y+\frac{1}{\sqrt{2}}$
Which is the required solution. So no option is correct.