= Solution
At an interior <local extremum> of a <differentiable function>, $f'(x)=0$. The <second-derivative test> classifies a <stationary point>: a positive <second derivative> gives a strict <local minimum>, and a negative <second derivative> gives a strict <local maximum>. A zero <second derivative> is inconclusive: $x^4$, $-x^4$, and $x^3$ at zero illustrate a minimum, a maximum, and neither.
Here
$$
f'(x)=3x^2-10x+3=(3x-1)(x-3),\qquad f''(x)=6x-10.
$$
Thus $f''(1/3)=-8$ and $f''(3)=8$. Substitution into $f$ gives
$$
\boxed{(1/3,-14/27)\text{ is a local maximum},\qquad (3,-10)\text{ is a local minimum}.}
$$
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