#y= cos^2x- sin^2x#

#y = (cosx + sinx)(cosx - sinx)#

By the product rule:

#y' = (cosx - sinx)(cosx - sinx) + (cosx + sinx)(-sinx - cosx)#

#y' = cos^2x - 2sinxcosx + sin^2x + (-cosxsinx - sin^2x - cos^2x - cosxsinx)#

#y' = cos^2x + sin^2x- 2sinxcosx + (-(cosxsinx + sin^2x + cos^2x + cosxsinx))#

#y' = 1 - 2sinxcosx + (-(2cosxsinx + 1))#

#y' = 1 - 2sinxcosx - 2sinxcosx - 1#

#y' = -4sinxcosx#

#y' = -2(2sinxcosx)#

#y' = -2sin2x#

The function will be increasing when #y' > 0# and decreasing when #y < 0#.

We set the derivative to #0# and solve for #x# first and then use test points to determine the intervals.

#0 = -2sin2x#

#0 = -4sinxcosx#

#x = 0, pi/2, pi, (3pi)/2, 2pi#

We try #pi/4, (3pi)/4, (5pi)/4, (7pi)/4# as test points. The results are as follows, with the function being #f(x)# and the derivative #f'(x)#

#f'(pi/4) = -4sin(pi/4)cos(pi/4) = -#

#f'((3pi)/4) = -4sin((3pi)/4)cos((3pi)/4) = +#

#f'((5pi)/4) = -4sin((5pi)/4)cos((5pi)/4) = -#

#f'((7pi)/4)) = -4sin((7pi)/4)cos((7pi)/4) = +#

Hence, in the interval #0 ≤ x ≤ 2pi#, the intervals of increase/decrease are:

•Decreasing over #0 ≤ x ≤ pi/2# and #pi ≤ x ≤ (3pi)/2#.

•Increasing over #pi/2 ≤ x ≤ pi# and #(3pi)/2 ≤ x ≤ 2pi#

Hopefully this helps!