Answer: \(\dfrac{dy}{dx}=9(3x-5)^2-2\), \(\dfrac{d^2y}{dx^2}=54(3x-5)\); \(x=\dfrac53\pm\dfrac{\sqrt2}{9}\); maximum at \(x=\dfrac53-\dfrac{\sqrt2}{9}\), minimum at \(x=\dfrac53+\dfrac{\sqrt2}{9}\).

Differentiate using the chain rule:

\(\frac{dy}{dx}=3(3x-5)^2\cdot3-2=9(3x-5)^2-2.\)

Differentiating again,

\(\frac{d^2y}{dx^2}=54(3x-5).\)

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

\(9(3x-5)^2-2=0.\)

Hence

\((3x-5)^2=\frac29,\)

so

\(3x-5=\pm\frac{\sqrt2}{3}.\)

Therefore

\(\boxed{x=\frac53\pm\frac{\sqrt2}{9}}.\)

For \(x=\frac53+\frac{\sqrt2}{9}\), \(3x-5\gt 0\), so \(\frac{d^2y}{dx^2}\gt 0\). This point is a minimum.

For \(x=\frac53-\frac{\sqrt2}{9}\), \(3x-5\lt 0\), so \(\frac{d^2y}{dx^2}\lt 0\). This point is a maximum.

Thus

\(\boxed{x=\frac53+\frac{\sqrt2}{9}\text{ is a minimum, and }x=\frac53-\frac{\sqrt2}{9}\text{ is a maximum}}.\)