$[x \sin ^2(\frac{y}{x})-y]dx+xdy=0$
$xdy=-(x \sin ^2(y/x)-y)dx$
=>$\large\frac{dy}{dx}=-\frac{(x \sin ^2(y/x)-y)}{x}$
$F(x,y)=\large\frac{-(x \sin ^2(y/x)-y)}{x}$
$F(tx,ty)=\large\frac{-(tx \; \sin ^2(ty/tx)-ty)}{tx}$
$=t^0 F(x,y)$
Hence $F(x,y)$ is a homogenous function of zero degree.
Now Put $y=vx$
$\large\frac{dy}{dx}=v+x \frac{dv}{dx}$
Therefore $v+x \large\frac{dv}{dx}=\frac{-(x \sin ^2v-vx)}{x}$
Therefore $v+x \large\frac{dv}{dx}=-\sin ^2v+v$
$=>x\; \large \frac{dv}{dx}=-\sin ^2v+v-v$
$=>x \;\large \frac{dv}{dx}=-\sin ^2v$
=>$\large\frac{dv}{\sin ^2 v}=-\frac{dx}{x}$
$=>cosec ^2 v dv=\large\frac{-dx}{x}$
integrating on both sides,
$\int cosec ^2 v\; dv=-\int \large\frac{dx}{x}$
$-\cot v=-log x-\log c$
Now substituting for V
$-\cot v=-log x - \log c$
Now substituting for V
$-\cot (y/x)=-(log cx)$
$\cot (y/x)=log cx$
$e^{\cot (y/x)}=cx$
When $y=\pi/4\; and\; x=1$
$\cot \pi/4=1 \qquad e^1=c$
Therefore $e=1 \qquad Therefore \; c=e$
Hence $e^{\cot (y/x)}=ex$
$=e^{\cot (x/y)-1}=x$