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The problem asks to calculate and classify the critical points of the function \( z = 4xy - 2x^2 - y^4 \).

### Step 1: Find the partial derivatives
To find the critical points, we need to calculate the first partial derivatives of the function with respect to \(x\) and \(y\).

1. Partial derivative with respect to \(x\):
   \[
   \frac{\partial z}{\partial x} = 4y - 4x
   \]

2. Partial derivative with respect to \(y\):
   \[
   \frac{\partial z}{\partial y} = 4x - 4y^3
   \]

### Step 2: Solve the system of equations
Set both partial derivatives equal to zero to find the critical points:

1. \( 4y - 4x = 0 \) ⟹ \( y = x \)
2. \( 4x - 4y^3 = 0 \) ⟹ \( x = y^3 \)

Substitute \( y = x \) into \( x = y^3 \):

\[
x = x^3
\]

Solve this equation:

\[
x(x^2 - 1) = 0
\]

Thus, \( x = 0 \) or \( x = \pm 1 \).

For each case:
- When \( x = 0 \), \( y = 0 \).
- When \( x = 1 \), \( y = 1 \).
- When \( x = -1 \), \( y = -1 \).

So, the critical points are: 
\[
(0, 0), (1, 1), (-1, -1)
\]

### Step 3: Classify the critical points
To classify the critical points, we use the second partial derivatives to calculate the Hessian matrix:

1. \( \frac{\partial^2 z}{\partial x^2} = -4 \)
2. \( \frac{\partial^2 z}{\partial y^2} = -12y^2 \)
3. \( \frac{\partial^2 z}{\partial x \partial y} = 4 \)

The Hessian matrix \( H \) is:
\[
H = \begin{pmatrix} \frac{\partial^2 z}{\partial x^2} & \frac{\partial^2 z}{\partial x \partial y} \\ \frac{\partial^2 z}{\partial x \partial y} & \frac{\partial^2 z}{\partial y^2} \end{pmatrix} = \begin{pmatrix} -4 & 4 \\ 4 & -12y^2 \end{pmatrix}
\]

Now, evaluate the determinant of the Hessian matrix \( D = \det(H) \) at each critical point:

1. **At \( (0, 0) \):**
   \[
   H = \begin{pmatrix} -4 & 4 \\ 4 & 0 \end{pmatrix}, \quad D = (-4)(0) - (4)(4) = -16
   \]
   Since \( D < 0 \), the critical point \( (0, 0) \) is a **saddle point**.

2. **At \( (1, 1) \):**
   \[
   H = \begin{pmatrix} -4 & 4 \\ 4 & -12 \end{pmatrix}, \quad D = (-4)(-12) - (4)(4) = 48 - 16 = 32
   \]
   Since \( D > 0 \) and \( \frac{\partial^2 z}{\partial x^2} = -4 < 0 \), the critical point \( (1, 1) \) is a **local maximum**.

3. **At \( (-1, -1) \):**
   \[
   H = \begin{pmatrix} -4 & 4 \\ 4 & -12 \end{pmatrix}, \quad D = 32
   \]
   Similarly, since \( D > 0 \) and \( \frac{\partial^2 z}{\partial x^2} = -4 < 0 \), the critical point \( (-1, -1) \) is a **local maximum**.

### Conclusion:
The critical points and their classifications are:
- \( (0, 0) \) is a saddle point.
- \( (1, 1) \) is a local maximum.
- \( (-1, -1) \) is a local maximum.

Would you like further details or clarification on any step? Here are some related questions for further exploration:

1. How is the Hessian determinant used to classify critical points?
2. What is the geometric interpretation of saddle points and local maxima?
3. Can there be other types of critical points for different functions?
4. How would the classification change if the function had more variables?
5. Why does \( D > 0 \) and \( \frac{\partial^2 z}{\partial x^2} < 0 \) indicate a local maximum?

**Tip**: Always remember to compute both the determinant and the sign of the second derivatives when classifying critical points using the Hessian method.

Calcule y clasifique los puntos críticos de z = 4xy - 2x^2 - y^4.

Calculate and classify the critical points of the function z = 4xy - 2x^2 - y^4 using multivariable calculus. This solution covers the partial derivatives, Hessian matrix, and classification of critical points as saddle points or local maxima.

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