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The problem asks to show that the partial derivatives \( f_{xy}(0, 0) \) and \( f_{yx}(0, 0) \) are not equal for the given function, defined piecewise as:

\[
f(x, y) = 
\begin{cases} 
xy & \text{if } |x| \geq |y| \\
-xy & \text{if } |x| < |y|
\end{cases}
\]

Let's proceed step by step to compute the mixed partial derivatives \( f_{xy}(0, 0) \) and \( f_{yx}(0, 0) \) and show that they are not equal.

### Step 1: Compute the Partial Derivatives

We begin by calculating the first-order partial derivatives of the function with respect to \(x\) and \(y\), and then the second-order mixed partial derivatives at the point \( (0, 0) \).

#### Case 1: \( |x| \geq |y| \)

Here, the function is \( f(x, y) = xy \).

1. The partial derivative with respect to \( x \):
   \[
   f_x(x, y) = y
   \]
2. The partial derivative with respect to \( y \):
   \[
   f_y(x, y) = x
   \]

#### Case 2: \( |x| < |y| \)

Here, the function is \( f(x, y) = -xy \).

1. The partial derivative with respect to \( x \):
   \[
   f_x(x, y) = -y
   \]
2. The partial derivative with respect to \( y \):
   \[
   f_y(x, y) = -x
   \]

### Step 2: Compute Mixed Partial Derivatives

Now, we compute the second-order mixed partial derivatives.

#### For \( f_{xy}(0, 0) \):

To compute \( f_{xy}(0, 0) \), we take the derivative of \( f_x(x, y) \) with respect to \( y \).

1. For \( |x| \geq |y| \), \( f_x(x, y) = y \), so:
   \[
   f_{xy}(x, y) = \frac{\partial}{\partial y} (y) = 1
   \]
2. For \( |x| < |y| \), \( f_x(x, y) = -y \), so:
   \[
   f_{xy}(x, y) = \frac{\partial}{\partial y} (-y) = -1
   \]

At \( (0, 0) \), both cases overlap, but the limit behavior in neighborhoods where \( |x| \geq |y| \) and \( |x| < |y| \) shows that the value of \( f_{xy}(0, 0) \) is dependent on the direction of approach. Therefore, the behavior is non-continuous, making \( f_{xy}(0, 0) \) undefined or non-equal in each region.

#### For \( f_{yx}(0, 0) \):

Similarly, to compute \( f_{yx}(0, 0) \), we take the derivative of \( f_y(x, y) \) with respect to \( x \).

1. For \( |x| \geq |y| \), \( f_y(x, y) = x \), so:
   \[
   f_{yx}(x, y) = \frac{\partial}{\partial x} (x) = 1
   \]
2. For \( |x| < |y| \), \( f_y(x, y) = -x \), so:
   \[
   f_{yx}(x, y) = \frac{\partial}{\partial x} (-x) = -1
   \]

Again, at \( (0, 0) \), the value of \( f_{yx}(0, 0) \) is dependent on the region of approach, and we observe non-continuous behavior, similar to the case of \( f_{xy} \).

### Conclusion:

The key observation is that the mixed partial derivatives \( f_{xy}(0, 0) \) and \( f_{yx}(0, 0) \) depend on the direction of approach due to the piecewise nature of the function. Therefore, they are not equal at \( (0, 0) \).

Would you like further details or have any questions on this? Below are some related questions that might help expand your understanding:

1. What are the formal conditions for equality of mixed partial derivatives?
2. How does the definition of differentiability relate to mixed partial derivatives?
3. Can we find other functions where \( f_{xy} \neq f_{yx} \)?
4. What is the geometric interpretation of mixed partial derivatives in 3D space?
5. How would this analysis change if the function were defined differently around \( (0, 0) \)?

**Tip:** Always check the continuity of partial derivatives when dealing with piecewise functions to understand their behavior at boundary points.

Show that f_xy(0, 0) ≠ f_yx(0, 0) for the function defined as f(x, y) = { xy if |x| ≥ |y|, -xy if |x| < |y| }.

Explore how to show that the mixed partial derivatives f_xy(0, 0) and f_yx(0, 0) are not equal for a piecewise function defined by conditions on |x| and |y|. This problem involves key multivariable calculus concepts such as partial derivatives and the application of Clairaut's Theorem. Suitable for college-level students.

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