To find the critical points of a #2#-variable function, you need to compute the gradient, which is a vector cointaining the derivatives with respect to each variable:
#(d/dx f(x,y), d/dy f(x,y))#
So, we have
#d/dx f(x,y) = 6cos(x)sin(y)#, and similarly
#d/dy f(x,y) = 6sin(x)cos(y)#.
In order to find the critical points, the gradient must be the zero vector #(0,0)#, which means solving the system
#{ (6cos(x)sin(y)=0), (6sin(x)cos(y)=0 ) :}#
which of course we can simplify getting rid of the #6#'s:
#{ (cos(x)sin(y)=0), (sin(x)cos(y)=0 ) :}#
This system is solved choosing for #x# a point which annihilates the cosine, and for #y# a point which annihilate the sine, and vice versa, so
#x=pm pi/2#, and #y=pm pi#, and vice versa #x=\pm pi# and #y=\pm pi/2#, obtaining #8# points in total.
