$\int_{0}^{\pi}\int_{0}^{\pi}\frac{1-\cos(3x+3y)}{2-\cos(x)-\cos(y)}dxdy$
After applying $\cos(C)+\cos(D)=2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$ this integral reduces to:
$$\int_{0}^{\pi}\int_{0}^{\pi}\frac{1-\cos(3x+3y)}{2\left(1-\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)\right)}dxdy$$
Now assuming that $x+y=t, x-y=u$, I get $x=(t+u)/2, y=(t-u)/2$, I got: $$\int\int\frac{1-\cos(3t)}{\left(1-\cos\left(\frac{t}{2}\right)\cos\left(\frac{u}{2}\right)\right)}dtdu$$
How to proceed further?
As I wrote in comments lines $x=0,x=\pi$ converted to lines $t+u=0,t+u=2\pi$ and lines $y=0,y=\pi$ converted to lines $t-u=0,t-u=2\pi$, so for integrals we have $$\int\limits_{0}^{\pi}\int\limits_{0}^{\pi}dxdy=\int\limits_{-\pi}^{0}\int\limits_{-u}^{2\pi+u}J\ dudt+\int\limits_{0}^{\pi}\int\limits_{u}^{2\pi-u}J\ dudt$$ Where Jakobian is $J$.