For this question we need to assume that a, b, c are constants, as opposed functions of x.
Secondly I believe the notation dy/dx f(x) is meant to mean df/dx, or alternatively f'(x).
1) f'(x) = 15ax² - 4bx + 4c
2) Given that the parabola y = 15ax² - 4bx + 4c >= 0
the vertex, in particular, must have a non-negative y coordinate.
The x-coordinate of the vertex is x = 2b/15a.
(If you do not know why that is, see NOTE 1 below.)
Thus the y-coordinate of the vertex is
y = 15a(2b/15a)² - 4b(2b/15a) + 4c = 4b²/15a - 8b²/15a + 4c
We can simplify the inequality
4b²/15a - 8b²/15a + 4c >= 0 (i.e., "vertex must have non-negative y coordinate")
by multiplying it by 15a.
We can simply multiply it by 15a because of the given assumption a > 0.
4b² - 8b² + 60ac >= 0
After algebraic simplification
b² <= 15ac
NOTE 1: Derivation of the formula for vertex.
Given a general parabola
y = Ax² + Bx + C
the vertex is where the derivative is 0:
2Ax + B = 0
x = -B/2A
NOTE 2: The assumption a > 0 did not have to be given to you; it is implied by f'(x) >= 0.
15ax² - 4bx + 4c >= 0
implies that the coefficient of x² must be positive, i.e.,
15a > 0