The solution of the differential equation $\dfrac{d y}{d x}+y \log y \cot x=0$ is

The given differential equation is $\dfrac{dy}{dx} + y \log y \cot x = 0$.

Rearranging the terms, we get $\dfrac{dy}{y \log y} = -\cot x dx$.

Integrating both sides, we have $\int \dfrac{dy}{y \log y} = -\int \cot x dx$.

Let $u = \log y$, then $du = \dfrac{1}{y} dy$. The integral becomes $\int \dfrac{du}{u} = -\int \cot x dx$.

This results in $\log |u| = -\log |\sin x| + C_1$.

Substituting $u = \log y$, we get $\log |\log y| = -\log |\sin x| + C_1$.

This simplifies to $\log |\log y| + \log |\sin x| = C_1$, which is $\log |\log y \sin x| = C_1$.

Taking the exponential of both sides, $|\log y \sin x| = e^{C_1} = C$.

Thus, $\sin x \log y = C$.

Answer: $\sin x \log y=c$