A curve has equation \(y=\dfrac{(3x^2-5)^{\frac13}}{x+4}\).
Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) can be written in the form \(\dfrac{Ax^2+Bx+C}{(3x^2-5)^{\frac23}(x+4)^2}\), where \(A\), \(B\) and \(C\) are integers.[5]
Hence find the \(x\)-coordinates of the stationary points on the curve. Give your answers in their simplest exact form.[3]