Answer: \(\dfrac{dy}{dx}=\dfrac{-x^2+8x+5}{(3x^2-5)^{2/3}(x+4)^2}\); stationary \(x\)-coordinates \(4-\sqrt{21}\) and \(4+\sqrt{21}\).

We start with the main method. Differentiate the expression and use the derivative to locate the required rate or stationary value.

The curve is

\(y=\frac{(3x^2-5)^{1/3}}{x+4}.\)

Use the quotient rule. The derivative of \((3x^2-5)^{1/3}\) is

\(\frac13(3x^2-5)^{-2/3}(6x)=2x(3x^2-5)^{-2/3}.\)

Therefore

\(\frac{dy}{dx}= \frac{2x(3x^2-5)^{-2/3}(x+4)-(3x^2-5)^{1/3}}{(x+4)^2}.\)

To combine the terms, factor out \((3x^2-5)^{-2/3}\) from the numerator:

\(\frac{dy}{dx}= \frac{(3x^2-5)^{-2/3}\{2x(x+4)-(3x^2-5)\}}{(x+4)^2}.\)

So

\(\frac{dy}{dx}= \frac{2x(x+4)-(3x^2-5)}{(3x^2-5)^{2/3}(x+4)^2}.\)

Simplify the numerator:

\(2x(x+4)-(3x^2-5)=2x^2+8x-3x^2+5=-x^2+8x+5.\)

Hence

\(\frac{dy}{dx}=\frac{-x^2+8x+5}{(3x^2-5)^{2/3}(x+4)^2}.\)

For stationary points, set the numerator equal to zero:

\(-x^2+8x+5=0.\)

So

\(x^2-8x-5=0.\)

Using the quadratic formula,

\(x=\frac{8\pm\sqrt{64+20}}{2}.\)

Therefore

\(x=4\pm\sqrt{21}.\)

This completes the solution and gives the required result.