Given curve $y = 8x^2 - x^4 -4$

Differentiate on both sides

$\frac{dy}{dx} = 16x - 4x^3$

To get stationary points $\frac{dy}{dx} = 0$

$4x\left(4-x^2\right)=0$

$\Rightarrow x = 0 , x^2 = 4$

$\Rightarrow x = 0 , \pm 2$

$x= 0 \Rightarrow y = 8\left(0\right)-0-4=-4$

$x = 2 \Rightarrow y = 8\left(4\right) - 16 - 4 = 12$

$x = -2 \Rightarrow y = 8\left(4\right) - 16 - 4 = 12$

So required stationary points are $\left(0, -4\right), \left(2, 12\right), \left(-2, 12\right).$