The equation of a curve is y = (2x − 3)3 − 6x.

(i) Express dy/dx and d^2y/dx^2 in terms of x

(ii) Find the x-coordinates of the two stationary points and determine the nature of each stationary
point.

y = (2x − 3)³ − 6x

(i)

dy/dx = 3 (2x − 3)² * 2 − 6
dy/dx = 6 (2x − 3)² − 6

d²y/dx² = 12 (2x − 3) * 2
d²y/dx² = 24 (2x − 3)

(ii)

Stationary points occur where dy/dx = 0

6 (2x − 3)² − 6 = 0
6 (2x − 3)² = 6
(2x − 3)² = 1
2x − 3 = ± 1
2x = 3 ± 1
x = (3 ± 1) / 2
x = 1, x = 2

When x = 1
d²y/dx² = 24 (2x − 3) = 6 (2−3) = −6 < 0
Therefore curve has local maximum at x = 1

When x = 2,
d²y/dx² = 24 (2x − 3) = 6 (4−3) = 6 > 0
Therefore curve has local minimum at x = 2

First of all, when you expand it, you see the function is simply y=-9

So the first derivative is 0, along with the second derivative.
0x

By stationary points I take it that they are critical points?
I don't see any changes in curvature at all.
Are you sure you typed it in right with no typos?