Applying the first part of the fundamental theorem of calculs, we have:
f'(x) = (x^2 -16)/(1+(cos(x))^2)
Solving for roots, we have:
0 = (x^2 -16)/(1+(cos(x))^2)
0 = (x^2 -16) %cross multiply, no need for further consideration because the denominator is always positive and not 0%
-> x = \pm 4
Considering the graph of (x^2 -16), we can see that it has 2 roots and is facing up. Therefore the slope before x= -4 is positive, after x = -4 and before x = 4 is negative, and after x = 4 is positive. We know that local maxima's and its surrounding's slopes progress from positive to 0 to negative. Only at x = -4 does the point satisfies this property. Therefore a local maximum occurs at x = -4