Solution:

The given function is f(x, y) = x^2 - 2y^2.

1. Find the gradient of the constraint equation:
The constraint equation is y - x^2 - 1 = 0. Taking the partial derivatives with respect to x and y, we get:
∂(y - x^2 - 1)/∂x = -2x
∂(y - x^2 - 1)/∂y = 1

2. Set up the Lagrangian function:
The Lagrangian function is given by L(x, y, λ) = f(x, y) + λ(g(x, y)), where g(x, y) is the constraint equation. In this case, g(x, y) = y - x^2 - 1. Thus, the Lagrangian function is:
L(x, y, λ) = x^2 - 2y^2 + λ(y - x^2 - 1)

3. Find the critical points:
To find the critical points, we need to solve the following system of equations:
∂L/∂x = 0
∂L/∂y = 0
g(x, y) = 0

Taking the partial derivatives of L(x, y, λ) with respect to x, y, and λ, we get:
∂L/∂x = 2x - 2λx = 0
∂L/∂y = -4y + λ = 0
g(x, y) = y - x^2 - 1 = 0

From the first equation, we can solve for λ in terms of x:
2x - 2λx = 0
2 - 2λ = 0
λ = 1

Substituting λ = 1 into the second equation, we get:
-4y + 1 = 0
y = 1/4

Substituting λ = 1 and y = 1/4 into the third equation, we get:
1/4 - x^2 - 1 = 0
x^2 = -3/4
This equation has no real solutions for x.

4. Determine the maximum value:
Since x does not have any real solutions, there are no critical points. Therefore, we cannot apply the Lagrange Multiplier method to find the maximum value of f(x, y) subject to the constraint.

However, we can observe that the given function f(x, y) = x^2 - 2y^2 is a quadratic function with a positive coefficient for the x^2 term. This means that the function opens upwards and has a minimum value, not a maximum value. Therefore, the statement that the maximum value of f(x, y) is 1 is incorrect.

In conclusion, the correct answer cannot be determined as the Lagrange Multiplier method cannot be applied to find the maximum value, and the given function does not have a maximum value.