Basically what we need to do is to derivate $ y(x,u)$ in terms of $ x$ and find $ u$ for which the derivative cancels. Before we do that, using the addition formula for tangents, we reduce the expression to the form \[ y = \frac 12 \cdot \frac { \cos { 2u} + 3 \cos {2x} + 2 }{ \sin (x-u) \sin (x+u) } .\] Now we can derivate easily, and obtain the equation \begin{align*} \cos 2 u \cot(u-x)+3 \cos 2x \cot(u-x)-\cos 2u\cot(u+x) & \\ -3 \cos 2x \cot (u+x)-6 \sin 2x & = 0. \end{align*} Via Wolfram, we get that the solutions we need are $ u = \dfrac 14 (4\pi k \pm \pi)$, where $ k\in \mathbb Z$.

i don't know if i'm mistaken, but i obtained that it was $ k\pi\pm\frac {\pi}{3}$, for integers $ k$.... i got also that $ y = \dfrac{3\cos 2x + \cos 2u + 2}{\cos 2u - \cos 2x} = - 3 + \dfrac{2 + 4\cos 2u}{\cos 2u - \cos 2x}$... so for the expression not to depend on $ x$ we should have that $ 2 + 4\cos 2u = 0$, which implies that $ u = k\pi\pm\frac {\pi}{3}$

60 degree.

I solved the given equation. y=(3cos^x - sin^u)/(sin^2x - sin^u) . Further i am not able simplify it for solution. Could some one please help me to simplify it. Pardon me for my ignorance as this is my first post and i don't know the rules of the forum. Thank you

campos, you're right, your approach is nicer than mine (I failed to see the denominator transformation). Seems your solution is correct. There might have been a typo with wolfram