Answer: The stationary points are \(\left(-\frac14,4\right)\) and \(\left(\frac14,4\right)\), and both are minimum points.

Differentiate first. Stationary points occur where the derivative is zero, while tangent and normal problems use the gradient at the given point.

Differentiate

\(y=32x^2+\frac{1}{8x^2} =32x^2+\frac18x^{-2}.\)

Then

\(\frac{dy}{dx}=64x-\frac14x^{-3} =64x-\frac{1}{4x^3}.\)

At a stationary point, \(\frac{dy}{dx}=0\), so

\(64x-\frac{1}{4x^3}=0.\)

Multiplying by \(4x^3\),

\(256x^4-1=0.\)

Therefore

\(x^4=\frac{1}{256}.\)

So

\(x=\pm\frac14.\)

For each of these values,

\(y=32\left(\frac{1}{16}\right)+\frac{1}{8(1/16)} =2+2=4.\)

Thus the stationary points are

For the second derivative,

\(\frac{d^2y}{dx^2}=64+\frac{3}{4x^4}.\)

Since \(x^4\gt 0\),

\(\frac{d^2y}{dx^2}\gt 0\)

at both stationary points. Therefore both stationary points are minimum points.