Solution of complex differential equation involving $2$ variables 
Solve the differential equation 
$\displaystyle \frac{dy}{dx} = \sqrt{\frac{1}{2}+\int^{-\sin^4 
 \theta}_{\cos^4 \theta}\frac{\sqrt{f(\phi)}d\phi}{\sqrt{f(\theta)}+\sqrt{f(\cos 2 \theta-\phi)}}},$
where $\theta = x+y$ and $\displaystyle x+y\in \bigg(\frac{\pi}{4}\;,\frac{3\pi}{4}\bigg)$

Try: First we will sove $$I = \int^{-\sin^4 
 \theta}_{\cos^4 \theta}\frac{\sqrt{f(\phi)}d\phi}{\sqrt{f(\theta)}+\sqrt{f(\cos 2 \theta-\phi)}}\cdots \cdots (1)$$
Replace $\phi = \cos^4 \theta-\sin^4 \theta-\phi = \cos (2\theta)-\phi.$
So $$I = \int^{-\sin^4 
 \theta}_{\cos^4 \theta}\frac{\sqrt{f(\cos 2 \theta -\phi)}d\phi}{\sqrt{f(\theta)}+\sqrt{f(\cos 2 \theta-\phi)}}\cdots (2)$$
So $$2I = \int^{-\sin^4 \theta}_{\cos^4 \theta}d\phi = \phi\bigg|^{-\sin^4 \theta}_{\cos^4 \theta}$$
$$I = \frac{1}{2}\bigg[\sin^4 \theta +\cos ^4 \theta\bigg] = \frac{1}{2}-\sin^2 \theta \cos^2 \theta$$
So $$\frac{dy}{dx} = \sqrt{1-\sin^2 \theta \cos^2 \theta}$$
could some help me how to solve further, i don,t get any clue
thanks
answer given as $\displaystyle \frac{2}{\sqrt{3}}\tan^{-1}\bigg(\frac{2\tan(x+y)+1}{\sqrt{3}}\bigg)=x+c.$

I'll assume that you meant to have $f(\phi)$ and not $f(\theta)$ in the denominator of the integral
You made a sign error
$$ I = \frac12 \left(-\sin^4 \theta - \cos^4 \theta \right) = -\frac12 +\sin^2\theta\cos^2\theta $$
Then
$$ \frac{dy}{dx} = \sqrt{\frac12 + I} = \big\vert\sin\theta\cos\theta\vert $$
You also have
$$ \frac{d\theta}{dx} = 1 + \frac{dy}{dx} = 1 + \big\vert\sin\theta\cos\theta\big\vert 
= \begin{cases} 
1 + \sin\theta\cos\theta, & \theta \in \left(\frac{\pi}{4}, \frac{\pi}{2} \right) \\
1 - \sin\theta\cos\theta, & \theta \in \left( \frac{\pi}{2}, \frac{3\pi}{4} \right)
 \end{cases} $$
Separating leads to the integral
$$ \int \frac{1}{1+\sin\theta\cos\theta}d\theta = \int \frac{\sec^2\theta}{\sec^2\theta+\tan\theta}d\theta = \int\frac{1}{u^2+u+1}du $$
where $u = \tan\theta$. Completing the square on the denominator gives
$$ \int\frac{1}{u^2+u+1}du = \int\frac{1}{\left(u+\frac12\right)^2+\frac34}du = \frac{2}{\sqrt{3}}\arctan\left(\frac{2u+1}{\sqrt{3}}\right) + c $$
