Discussion
For the function \(f(x, y) = (x - 1)(x^{2} - y^{2})\), the point \(\left(\dfrac{2}{3}, 0\right)\) :
Answer: C
The point \((2/3, 0)\) is a local minimum.
Expand \(f = (x - 1)(x^{2} - y^{2})\). Then \(f_{x} = 3x^{2} - 2x - y^{2}\) and \(f_{y} = -2y(x - 1)\). At \((2/3, 0)\): \(f_{x} = 4/3 - 4/3 = 0\) and \(f_{y} = 0\), so it is a critical point.
Second derivatives: \(f_{xx} = 6x - 2 = 2\), \(f_{yy} = -2(x - 1) = 2/3\), \(f_{xy} = -2y = 0\). So
\[ \begin{aligned} f_{xx}f_{yy} - f_{xy}^{2} &= 2 \cdot \tfrac{2}{3} - 0 \\ &= \tfrac{4}{3} \gt 0 \end{aligned} \]
with \(f_{xx} \gt 0\), which is a local minimum. The other critical points, (0, 0), (1, 1) and (1, -1), are where the saddle points of this function lie, which is the trap in option D.
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