Given: y = 2sin³(x) - 3sin(x)

i. First, let's find the first derivative dy/dx:

dy/dx = d/dx [2sin³(x) - 3sin(x)]
= 6sin²(x)cos(x) - 3cos(x)
= 3cos(x)[2sin²(x) - 1]

To find the stationary points, we need to solve dy/dx = 0:

3cos(x)[2sin²(x) - 1] = 0

Setting each factor equal to zero gives us two cases:

3cos(x) = 0:
This occurs when x = π/2 (90 degrees) and x = 3π/2 (270 degrees).

2sin²(x) - 1 = 0:
Rearranging this equation gives sin²(x) = 1/2.
Taking the square root, we have sin(x) = ±1/√2.
This occurs when x = π/4 (45 degrees) and x = 3π/4 (135 degrees).

Therefore, the coordinates of the stationary points are:
(π/2, y(π/2)), (3π/2, y(3π/2)), (π/4, y(π/4)), and (3π/4, y(3π/4)).

ii. Next, let's find the second derivative d²y/dx²:

d²y/dx² = d/dx [3cos(x)[2sin²(x) - 1]]
= -6sin(x)cos²(x) - 3sin(x)
= -3sin(x)[2cos²(x) + 1]

Now, we can apply the Second Derivative Test:

For x = π/2 (90 degrees):
Plugging x = π/2 into d²y/dx² gives:
-3sin(π/2)[2cos²(π/2) + 1]
-3(1)[2(0) + 1]
= -3

Since the second derivative is negative, the point (π/2, y(π/2)) is a local maximum.

For x = 3π/2 (270 degrees):
Plugging x = 3π/2 into d²y/dx² gives:
-3sin(3π/2)[2cos²(3π/2) + 1]
-3(-1)[2(0) + 1]
= 3

Since the second derivative is positive, the point (3π/2, y(3π/2)) is a local minimum.

For x = π/4 (45 degrees):
Plugging x = π/4 into d²y/dx² gives:
-3sin(π/4)[2cos²(π/4) + 1]
-3(√2/2)[2(√2/2)² + 1]
= -3(√2/2)[2(1/2) + 1]
= -3(√2/2)(2/2 + 1)
= -3(√2/2)(3/2)
= -9√2/4

Since the second derivative is negative, the point (π/4, y(π/4)) is a local maximum.

For x = 3π/4 (135 degrees):
Plugging x = 3π/4 into d²y/dx² gives:
-3sin(3π/4)[2cos²(3π/4) + 1]
-3(-√2/2)[2(-√2/2)² + 1]
= 3(√2/2)[2(1/2) + 1]
= 3(√2/2)(2/2 + 1)
= 3(√2/2)(3/2)
= 9√2/4

Since the second derivative is positive, the point (3π/4, y(3π/4)) is a local minimum.

Therefore, the nature of each stationary point is as follows:

(π/2, y(π/2)): Local maximum
(3π/2, y(3π/2)): Local minimum
(π/4, y(π/4)): Local maximum
(3π/4, y(3π/4)): Local minimum