I have a three-variable function for $(x,y)\in[-\pi,\pi]$ and $z>0$
$f_z (x,y)=A(z)(\cos x+\cos y)+B(z)(\cos 2x+\cos 2y)+C(z) (\cos x \cdot\cos y),$
in which the real functions $A(z),B(z),C(z)$ can be positive or negative. I am looking for those points $(x,y)$ at which the function $f_z (x,y)$ reaches its extrema (both minimum and maximum).
Question
I may probably find those points by using the Hessian method for obtaining the extrema. I am asking if there is any simpler/obvious way to find those points.
I assume you have a family of functions indexed by a parameter $z$,and wish to study for each fixed choice of $z$ what are the extrema of the function $f_z(x,y)$. Following up on Anne Bauval's comment, the quadratic function $g(u,v)$ may have interior critical points where (i) $g_u=a + 2 b u + cv=0$ and also (ii) $ g_v= a +2b v +cu =0$. (They are interior solutions if they lie in the region $-1<u<1$ and $-1<v<1$.)
Solving this pair of linear equations we get
$u =v = \frac{ a( c-2b)}{D}$ where in your problem $D= 4b^2-c^2$ is the discriminant. The classification of these interior critical points of $g$ involves the use of the Hessian matrix, specifically first checking the sign of the discriminant $D$ and then the sign of $b$.
(You also get another family of critical points where $0=D=4b^2=c^2$; these satisfy $ u=-v$. The Hessian test is indeterminant in this case, and more complicated tests are needed to classify these.
Here is a visual check on the location of critical points in a worked example. I plotted the solution surfaces of (i) and (ii) with 3D graphics and see where they intersect in space.
This is the intersection set of the two surfaces. Basically the set of critical points consists of three blue curves, Two curves lie in the plane where $u=v$ and one curve lies in the plane $u=-v$. Looking down from above you see they lie in two perpendicular planes.
(Note that two curves merge in a disaster at one place).