Find the local maximum and minimum values, and saddle point(s) of f(x,y). f(x,y)=y^3-3y+3x^2y.

To find the local maximum, local minimum, and saddle points of the function \( f(x, y) = y^3 - 3y + 3x^2y \), we follow these steps:

### 1. Find the Critical Points

First, we need to find the partial derivatives of \( f(x, y) \) and set them to zero.

#### Partial Derivatives:
\[ f_x = \frac{\partial f}{\partial x} = 6xy \]
\[ f_y = \frac{\partial f}{\partial y} = 3y^2 - 3 + 3x^2 \]

Set these partial derivatives equal to zero to find the critical points.

#### Solve \( f_x = 0 \):
\[ 6xy = 0 \]
This gives two possibilities:
1. \( x = 0 \)
2. \( y = 0 \)

#### Solve \( f_y = 0 \):
\[ 3y^2 - 3 + 3x^2 = 0 \]
\[ y^2 + x^2 - 1 = 0 \]
\[ y^2 = 1 - x^2 \]

Combine the conditions:

1. For \( x = 0 \):
   \[ y^2 = 1 \]
   \[ y = \pm 1 \]
   So, the critical points are \( (0, 1) \) and \( (0, -1) \).

2. For \( y = 0 \):
   \[ y^2 + x^2 - 1 = 0 \]
   \[ x^2 = 1 \]
   \[ x = \pm 1 \]
   So, the critical points are \( (1, 0) \) and \( (-1, 0) \).

### 2. Classify the Critical Points

We use the second derivative test, which involves computing the Hessian matrix.

#### Second Partial Derivatives:
\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6y \]
\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y \]
\[ f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = 6x \]

The Hessian matrix \( H \) is:
\[ H = \begin{pmatrix}
f_{xx} & f_{xy} \\
f_{xy} & f_{yy}
\end{pmatrix} = \begin{pmatrix}
6y & 6x \\
6x & 6y
\end{pmatrix} \]

The determinant of \( H \) is:
\[ \text{det}(H) = (6y)(6y) - (6x)(6x) = 36y^2 - 36x^2 = 36(y^2 - x^2) \]

Evaluate the determinant at each critical point:

1. At \( (0, 1) \):
   \[ \text{det}(H) = 36(1^2 - 0^2) = 36 > 0 \]
   \[ f_{xx} = 6 \cdot 1 = 6 > 0 \]
   So, \( (0, 1) \) is a local minimum.

2. At \( (0, -1) \):
   \[ \text{det}(H) = 36((-1)^2 - 0^2) = 36 > 0 \]
   \[ f_{xx} = 6 \cdot (-1) = -6 < 0 \]
   So, \( (0, -1) \) is a local maximum.

3. At \( (1, 0) \) and \( (-1, 0) \):
   \[ \text{det}(H) = 36(0^2 - 1^2) = -36 < 0 \]
   So, \( (1, 0) \) and \( (-1, 0) \) are saddle points.

### Summary

- **Local Maximum**: \( (0, -1) \)
- **Local Minimum**: \( (0, 1) \)
- **Saddle Points**: \( (1, 0) \) and \( (-1, 0) \)

Would you like more details or have any questions? 

Here are some related questions you might be interested in:
1. How do you find critical points for functions of multiple variables?
2. What is the Hessian matrix and how is it used?
3. Can you show examples of saddle points in different functions?
4. How do you determine if a critical point is a local minimum, maximum, or saddle point?
5. What are the applications of finding local maxima and minima in real-world problems?

**Tip:** When analyzing functions of several variables, always check the second partial derivatives and the determinant of the Hessian matrix to classify the critical points accurately.

Learn how to find local maximum, minimum, and saddle points for the function f(x, y) = y^3 - 3y + 3x^2y using partial derivatives and the second derivative test. This problem involves multivariable calculus concepts and applications of critical points. Suitable for advanced undergraduate students and researchers.

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