1) y = x (1-\(x^{2}\))^(\(\frac{1}{2}\))
y' = (1-\(x^{2}\))^(\(\frac{1}{2}\)) + x・2x(1-\(x^{2}\))^(-\(\frac{1}{2}\))/2
= (1-\(x^{2}\))^(\(\frac{1}{2}\)) + \(x^{2}\)(1-\(x^{2}\))^(-\(\frac{1}{2}\))
= ((1-\(x^{2}\)) + \(x^{2}\))/\(\sqrt{\quad}\)(1-\(x^{2}\))
= 1/\(\sqrt{\quad}\)(1-\(x^{2}\)) > 0, -1 ＜ x ＜ 1.
従って極値は存在しない。

2) ∂f/∂x = y + 2xy - \(y^{2}\) = y(1 + 2x - y),
∂f/∂y = x(1 + x - 2y).

ここで ∂f/∂x =∂f/∂y = 0 と置くと
x = y = 0
又は
y = 0 且つ 1 + x - 2y = 0 (即ち x = -1, y = 0)
又は
1 + 2x - y = 0 且つ x = 0 (即ち x = 0, y = 1)
又は
1 + 2x - y = 1 + x - 2y = 0 (即ち x = -\(\frac{1}{3}\), y = \(\frac{1}{3}\)).

∂^2f/∂\(x^{2}\) = 2y,
∂^2f/(∂x∂y) = 1 + 2x - 2y,
∂^2f/∂\(y^{2}\) = -2x.
よって Hessian
H(x, y) = 2y×(-2x) - (1 + 2x - 2y\()^{2}\)
= - 4xy - (4\(x^{2}\) - 4xy + 4\(y^{2}\) + 4x - 4y + 1)
= -4\(x^{2}\) -4\(y^{2}\) - 4x + 4y - 1.

H(0, 0) = -1 ＜ 0 は極値点ではない。
H(-1, 0) = -4 + 4 - 1 = -1 ＜ 0 も極値点ではない。
H(0, 1) = -4 + 4 -1 = -1 ＜ 0 も極値点ではない。
H(-\(\frac{1}{3}\), \(\frac{1}{3}\)) = \(\frac{7}{9}\) > 0 なのでこれが極値点。
極値は
f(-\(\frac{1}{3}\), \(\frac{1}{3}\)) = -\(\frac{1}{27}\).
(∂^2/∂\(x^{2}\))f(-\(\frac{1}{3}\), \(\frac{1}{3}\)) = \(\frac{2}{3}\) だから下に凸なので, 極小。

3)∂f/∂x = 4\(x^{3}\) - 12\(x^{2}\) + 6x - 2y
　= 2(2\(x^{3}\) - 6\(x^{2}\) + 3x - y), ∂f/∂y = -2(x + y).

ここで ∂f/∂x =∂f/∂y = 0 と置くと
y = -x だから
2\(x^{3}\) - 6\(x^{2}\) + 3x + x
= 2\(x^{3}\) - 6\(x^{2}\) + 4x
= 2x(\(x^{2}\) - 3x + 2)
= 2x(x - 1)(x - 2) = 0.
即ち (x, y) = (0, 0), (1, -1), (2, -2).

∂^2f/∂\(x^{2}\) = 6(2\(x^{2}\) - 4x + 1),
∂^2f/(∂x∂y) = -2,
∂^2f/∂\(y^{2}\) = -2.
よって Hessian
H(x, y) = -12(2\(x^{2}\) - 4x + 1) - 4 = -8(3\(x^{2}\) - 6x + 2).

H(0, 0) = -8 ＜ 0 は極値点ではない。
H(1, -1) = -8×(3 - 6 + 2) = 8 > 0 は極値点。
H(2, -2) = -8×(12 - 12 + 2) = -16 ＜ 0 は極値点ではない。
極値は
f(1, -1) = 1 - 1 + 3 + 2 - 1 = 2.
(∂^2/∂\(x^{2}\))f(1, -1) = 6×(2 - 4 + 1) = -6 ＜ 0 は上に凸なので極大。