if cos x

$\cos x d y-(\sin x) y d x=6 x d x$

$\Rightarrow \int d(y \cos x)=\int 6 x d x \Rightarrow y \cos x=3 x^{2}+C \ldots(1)$

Given, $y\left(\frac{\pi}{3}\right)=0$

Putting $x=\frac{\pi}{3}$ and $y=0$ in eq. (1), we get

$(10) \times\left(\frac{1}{2}\right)=\frac{3 \pi^{2}}{9}+C \Rightarrow C \frac{-\pi^{2}}{3}$

So, from (1) $y \cos x=3 x^{2}-\frac{\pi^{2}}{3}$

Now, put $\quad-$ in the above equation,

$y \frac{\sqrt{3}}{2}=\frac{3 \pi^{2}}{36}-\frac{\pi^{2}}{3} \Rightarrow \frac{\sqrt{3} y}{2}=\frac{-3 \pi^{2}}{12} \Rightarrow y=\frac{-\pi^{2}}{2 \sqrt{3}}$