Answer: The only stationary point is \((0,32)\), and it is a minimum point.

Differentiate using the chain rule:

\(\frac{dy}{dx} = \frac53(3x^2+8)^{2/3}\cdot 6x.\)

So

\(\frac{dy}{dx}=10x(3x^2+8)^{2/3}.\)

For a stationary point,

\(10x(3x^2+8)^{2/3}=0.\)

This gives \(x=0\), since \(3x^2+8=0\) has no real solution.

Therefore there is only one stationary point.

When \(x=0\),

\(y=(8)^{5/3}=(2^3)^{5/3}=2^5=32.\)

So the stationary point is

\((0,32).\)

Since \((3x^2+8)^{2/3}\) is always positive, the sign of \(\dfrac{dy}{dx}\) is the sign of \(x\).

So \(\dfrac{dy}{dx}\lt 0\) when \(x\lt 0\), and \(\dfrac{dy}{dx}\gt 0\) when \(x\gt 0\).

The curve changes from decreasing to increasing, so \((0,32)\) is a minimum point.