Introduction:
To prove that the function f(x, y) = x^2 - 2xy + y^2 - x^3 + y^3 - x^5 does not have a maximum or minimum at the origin (0, 0), we can use the Second Partial Derivative Test. This test involves finding the second partial derivatives of the function and evaluating them at the critical point (0, 0) to determine the nature of the critical point.

Second Partial Derivative Test:
The Second Partial Derivative Test states that if the second partial derivatives of a function f(x, y) exist and satisfy the following conditions at a critical point (a, b):
1. The determinant of the Hessian matrix, D = f_xx(a, b) * f_yy(a, b) - f_xy(a, b)^2, is positive,
2. f_xx(a, b) > 0,

Then the critical point (a, b) is a local minimum.

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If the conditions are not satisfied, the nature of the critical point cannot be determined using this test.

Evaluating the Second Partial Derivatives:
To apply the Second Partial Derivative Test, we need to find the second partial derivatives of f(x, y) = x^2 - 2xy + y^2 - x^3 + y^3 - x^5.

Taking the first partial derivative with respect to x, we get:
f_x = 2x - 2y - 3x^2 - 5x^4

Taking the first partial derivative with respect to y, we get:
f_y = -2x + 2y + 3y^2

Taking the second partial derivative with respect to x, we get:
f_xx = 2 - 6x - 20x^3

Taking the second partial derivative with respect to y, we get:
f_yy = 2 + 6y

Taking the mixed partial derivative with respect to x and y, we get:
f_xy = -2

Evaluating the Second Partial Derivatives at the Critical Point:
To evaluate the second partial derivatives at the critical point (0, 0), we substitute x = 0 and y = 0 into the expressions for f_xx, f_yy, and f_xy.

f_xx(0, 0) = 2 - 6(0) - 20(0)^3 = 2
f_yy(0, 0) = 2 + 6(0) = 2
f_xy(0, 0) = -2

Determining the Nature of the Critical Point:
Now, we can determine the nature of the critical point (0, 0) by evaluating the determinant D = f_xx(0, 0) * f_yy(0, 0) - f_xy(0, 0)^2.

D = 2 * 2 - (-2)^2 = 4 - 4 = 0

Since the determinant D is zero, the Second Partial Derivative Test cannot determine the nature of the critical point (0, 0). Therefore, the function f