since cos(x) + cos(x+y) = 0

and since x = y

cos(x) + cos(2x) = 0

2cos^2(x) +cos(x) -1 = 0

(2cos(x) -1)(cos(x)+1) = 0

cos(x) = 1/2 cos(x) = -1

I'm assuming the domain is the square [0,2pi] x[0,2pi] ?

In which case x = pi/3, 5pi/3, pi

Again since x = y

(pi/3,pi/3) , (5pi/3,5pi/3) (pi,pi) are the critical points