Given value curve is
$\sin y=x \sin \left(\frac{\pi}{3}+y\right)$
$\Rightarrow \cos y \frac{d y}{d x}=x \cos \left(\frac{\pi}{3}+y\right) \frac{d y}{d x}+\sin \left(\frac{\pi}{3}+y\right)$
$\Rightarrow \left(\frac{d y}{d x}\right)_{x=0}=\frac{\sqrt{3}}{2}$
$\Rightarrow \left(\frac{-d x}{d y}\right)_{x=0}=\frac{-2}{\sqrt{3}}$
$\therefore $ Equation of normal at $(0,0)$ is $(y-0)=\frac{-2}{\sqrt{3}}(x-0)$
$\Rightarrow y=\frac{-2}{\sqrt{3}} x$ or $2 x+\sqrt{3} y=0$