On differentiating we get
\(\frac{1}{y}\) dy = dt
\( \frac{dt}{t}\) = - cot x dx
log t = -log (sin x) + C
log t + log(sin x) = C
log(tsin x) = C
tsin x = C
(log y)(sin x) = C
Conclusion: Therefore, (log y)(sin x) = C is the solution of \(\frac{dy}{dx}\) + y log y cot x = 0