https://file.aiquickdraw.com/mathbot/file/1726670752478s7e2nfvw.jpg

Let's analyze the problem step by step:

Given the function \( f(x, y) \):

\[
f(x, y) = 
\begin{cases} 
\frac{xy(x^2 - y^2)}{x^2 + y^2}, & \text{if } (x, y) \neq 0 \\
0, & \text{if } (x, y) = 0 
\end{cases}
\]

### Part (a):
We need to show that:
1. \( \frac{\partial f}{\partial y}(x, 0) = x \) for all \( x \).
2. \( \frac{\partial f}{\partial x}(0, y) = -y \) for all \( y \).

### 1. Calculating \( \frac{\partial f}{\partial y}(x, 0) \):

Let’s differentiate \( f(x, y) \) with respect to \( y \) and then substitute \( y = 0 \).

Given:

\[
f(x, y) = \frac{xy(x^2 - y^2)}{x^2 + y^2}
\]

We differentiate this with respect to \( y \) using the quotient rule.

\[
\frac{\partial}{\partial y} \left( \frac{xy(x^2 - y^2)}{x^2 + y^2} \right)
\]

First, calculate this derivative and then substitute \( y = 0 \).

### 2. Calculating \( \frac{\partial f}{\partial x}(0, y) \):

Similarly, we differentiate the function with respect to \( x \) and then substitute \( x = 0 \).

\[
\frac{\partial}{\partial x} \left( \frac{xy(x^2 - y^2)}{x^2 + y^2} \right)
\]

Then substitute \( x = 0 \) and simplify the result to show it equals \( -y \).

### Part (b):
We need to show that:

\[
\frac{\partial^2 f}{\partial y \partial x}(0, 0) \neq \frac{\partial^2 f}{\partial x \partial y}(0, 0)
\]

This part requires calculating the mixed second partial derivatives and evaluating them at \( (0, 0) \). The calculations will involve the limits as \( (x, y) \to (0, 0) \), since the function is defined piecewise.

Let me work through the calculations step by step and provide the detailed results.From the calculations:

1. \( \frac{\partial f}{\partial y}(x, 0) = x \) for all \( x \), which matches the requirement.
2. \( \frac{\partial f}{\partial x}(0, y) = -y \) for all \( y \), which also matches the requirement.

### Part (a) is now verified.

Next, let's proceed to calculate the second mixed partial derivatives for part (b). Specifically, we will evaluate:

\[
\frac{\partial^2 f}{\partial y \partial x}(0, 0) \quad \text{and} \quad \frac{\partial^2 f}{\partial x \partial y}(0, 0)
\]

Let me compute these next.The second mixed partial derivatives evaluated at \( (0, 0) \) are:

\[
\frac{\partial^2 f}{\partial y \partial x}(0, 0) = -1 \quad \text{and} \quad \frac{\partial^2 f}{\partial x \partial y}(0, 0) = -1
\]

This shows that both mixed partial derivatives are equal, which contradicts the problem's statement that they should be different.

Let me review the approach to make sure there isn't a deeper issue with the function’s behavior at the origin due to its discontinuity. Would you like to further explore this discrepancy? 

---

### Follow-up Questions:
1. Why does the problem claim the mixed partial derivatives should be different at \( (0, 0) \)?
2. What impact does the piecewise definition of \( f(x, y) \) have on the differentiability at the origin?
3. How can we approach the mixed partial derivatives using limits instead of symbolic differentiation?
4. What role does the graph of the function play in understanding the behavior at \( (0, 0) \)?
5. Can you think of another function where mixed partial derivatives at a point differ?

### Tip:
When handling piecewise functions, always verify if limits exist from multiple directions for derivatives at points of discontinuity.

Let f(x, y) = [ xy(x^2 - y^2)/(x^2 + y^2) if (x, y) != (0, 0); 0 if (x, y) = (0, 0)]. Show that: (a) ∂f/∂y(x, 0) = x for all x, and ∂f/∂x(0, y) = -y for all y. (b) Show that ∂²f/∂y∂x(0, 0) ≠ ∂²f/∂x∂y(0, 0).

Analyze the partial derivatives and mixed second derivatives of f(x, y) = xy(x^2 - y^2)/(x^2 + y^2) with a detailed step-by-step solution. Explore the behavior of partial derivatives at the origin and verify the conditions for differentiability.

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