Find the first derivative, (dy)/(dx).

$$4x^3 - 32x = 0$$

Factor out $4x$:

$$4x(x^2 - 8) = 0$$

This gives three possible values for $x$:

$$4x = 0 \implies x = 0$$ $$x^2 - 8 = 0 \implies x^2 = 8 \implies x = \pm\sqrt{8} = \pm 2\sqrt{2}$$

The stationary points occur at $x = 0$, $x = 2\sqrt{2}$, and $x = -2\sqrt{2}$.

For $x = 0$:
Substitute $x=0$ into the second derivative:

$$\frac{d^2y}{dx^2} \Big|_{x=0} = 12(0)^2 - 32 = -32$$

Since $-32 < 0$, the stationary point at $x = 0$ is a local maximum.

For $x = 2\sqrt{2}$:
Substitute $x=2\sqrt{2}$ into the second derivative:

$$\frac{d^2y}{dx^2} \Big|_{x=2\sqrt{2}} = 12(2\sqrt{2})^2 - 32$$ $$= 12(4 \times 2) - 32$$ $$= 12(8) - 32$$ $$= 96 - 32 = 64$$

Since $64 > 0$, the stationary point at $x = 2\sqrt{2}$ is a local minimum.

For $x = -2\sqrt{2}$:
Substitute $x=-2\sqrt{2}$ into the second derivative:

$$\frac{d^2y}{dx^2} \Big|_{x=-2\sqrt{2}} = 12(-2\sqrt{2})^2 - 32$$ $$= 12(4 \times 2) - 32$$ $$= 12(8) - 32$$ $$= 96 - 32 = 64$$

Since $64 > 0$, the stationary point at $x = -2\sqrt{2}$ is a local minimum.