Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 42

You need to evaluate the monotony of the function, hence, you need to remember that the function increases if f'(x)>0 and the function decreases if f'(x)<0.
You need to evaluate the first derivative of the function:
f'(x) = (ln(x^4+27))'
f'(x) = (1/(x^4+27))*(x^4+27)'
f'(x) = (4x^3)/(x^4+27)
You need to set f'(x) = 0:
(4x^3)/(x^4+27) = 0
4x^3 = 0 => x = 0
You need to notice that f'(x)>0 for x in (0,+oo) and f'(x)<0 for x in (-oo,0), hence, the function increases for x in (0,+oo) and it decreases for x in (-oo,0).
b) The local maximum and minimum values are those x values for f'(x) = 0. From previous point a) yields that f'(x) = 0 for x = 0 and the function decreases as x approaches to 0, from the left, and then it increases.
Hence, the function has only a minimum point at x = 0, and the point is (0, ln27).