<s>[INST] My attempt regarding finding critical ponts of $(\cos x)(\cos y)(\cos(x+y))$ Given this problem Restrictions on $x$ any are that $x\in[0,\pi]$ , $y\in[0,\pi]$
I have $f_x=-(\cos y)({\sin(2x+y))}--------*$  
$f_y=-(\cos x)(\sin x+2y)-----------**$
So from $*$ I get either $\cos y=0$ or $\sin(2x+y)=0$
From $\sin(2x+y)=0$ , I get $2x+y=0,\pi,2\pi,3\pi$ as $2x+y\in[0,3\pi]$
And from second case I get $\cos y=0\implies y=\dfrac{\pi}{2}$
Also from $**$ , I get $x+2y=0,\pi,2\pi,3\pi$ and from $\cos x=0\implies x=\dfrac{\pi}{2}$
Now I have made $16$ cases because of following equations:
$2x+y=0,\pi,2\pi,3\pi$ 
$x+2y=0,\pi,2\pi,3\pi$
example as like $2x+y=0$ , $x+2y=3\pi$ ....similarly $16$ cases from above equations
and I point $\left(\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
After examining all cases I have got valid critical points only for cases of type  
$2x+y=R$
$x+2y=R$
Where $R$ is $0,\pi,2\pi,3\pi$ , taken one by one. So out of $16$ cases I have made above , I got my valid points out of $4$ cases as shown above..
MY QUESTION - Why I am getting points from these cases only ($R$ cases) ...Points from other $12$ cases are either same as that of the ($R$ cases), or they are not in domain 
Also am I right in making $16$ cases like this. Is this correct way of solving these equations? 
Is there any alternative way??
Thanks for your kind help!!
 [/INST] You have $f(x,y)=\cos x\cos y\cos(x+y)$, so
$$
f'_x=-\sin x\cos y\cos(x+y)-\cos x\cos y\sin(x+y)=-\cos y\sin(2x+y)
$$
Since interchanging $x$ and $y$ doesn't change the function, we have
$$
f'_y=-\cos x\sin(x+2y)
$$
We have four cases:
$$
\begin{cases}
\cos y=0\\
\cos x=0
\end{cases}
\qquad
\begin{cases}
\cos y=0\\
\sin(x+2y)=0
\end{cases}
\qquad
\begin{cases}
\sin(2x+y)=0\\
\cos x=0
\end{cases}
\qquad
\begin{cases}
\sin(2x+y)=0\\
\sin(x+2y)=0
\end{cases}
$$
The first case gives $x=y=\pi/2$.
The second case gives $y=\pi/2$ and $\sin(x+\pi)=0$, that is, $-\sin x=0$, so critical points at $(0,\pi/2)$ and $(\pi,\pi/2)$.
The third case is symmetric, giving critical points at $(\pi/2,0)$ and $(\pi/2,\pi)$.
Let's tackle the last case. From $\sin(2x+y)=0$ we get $2x+y=m\pi$ for some integer $m$; similarly, $x+2y=n\pi$ for an integer $n$. Now let's solve the linear system
$$
\begin{cases}
2x+y=m\pi\\
x+2y=n\pi
\end{cases}
$$
that gives
$$
\begin{cases}
x=(2m-n)\dfrac{\pi}{3}\\[6px]
y=(2n-m)\dfrac{\pi}{3}
\end{cases}
$$
With the given limitations we must have $0\le 2m-n\le 3$ and $0\le 2n-m\le 3$ and we should determine $m$ and $n$ so that this holds.
If we set $2m-n=h$ and $2n-m=k$, we get
$$
m=\frac{2h+k}{3},\qquad n=\frac{h+2k}{3}
$$
So we cannot have arbitrary choices for $h$ and $k$, since we want integer values of $m$ and $n$. The only values are so
\begin{alignat}{3}
&m=0, n=0&\quad&\text{corresponding to}&\quad& h=0, k=0
\\
&m=1, n=1,&&\text{corresponding to}&& h=1, k=1
\\
&m=1, n=2&&\text{corresponding to}&& h=0, k=3
\\
&m=2, n=1,&&\text{corresponding to}&& h=3, k=0
\\
&m=2, n=2,&&\text{corresponding to}&& h=2, k=2
\\
&m=3, n=3&&\text{corresponding to}&& h=3, k=3
\end{alignat}
and the critical points are
$$
\left(0,0\right)
\qquad
\left(\frac{\pi}{3},\frac{\pi}{3}\right)
\qquad
\left(0,\pi\right)
\qquad
\left(\pi,0\right)
\qquad
\left(\frac{2\pi}{3},\frac{2\pi}{3}\right)
\qquad
\left(\pi,\pi\right)
$$
See the following diagram drawn with Geogebra to see why

</s>