\[\frac{dy}{dx}=3x^2-6x=3x(x-2).\] Stationary points have \(x=0\) or \(x=2\), giving \((0,9)\) and \((2,5)\). Also \(d^2y/dx^2=6x-6\). At \(x=0\), this is \(-6<0\); at \(x=2\), it is \(6>0\). Thus \(\boxed{(0,9)\text{ is a maximum,}\quad(2,5)\text{ is a minimum}}\).