u = log(x^3+y^3-x^2y-xy^2) show that Uxx+2Uxy+Uyy = -4/(x+y)^2

Given the function \( u = \log(x^3 + y^3 - x^2y - xy^2) \), we are asked to show that:

\[
u_{xx} + 2u_{xy} + u_{yy} = -\frac{4}{(x + y)^2}
\]

### Step 1: Compute the First Derivatives \( u_x \) and \( u_y \)

First, compute the partial derivatives of \( u \) with respect to \( x \) and \( y \):

\[
u_x = \frac{\partial u}{\partial x} = \frac{1}{f(x, y)} \cdot \frac{\partial f(x, y)}{\partial x}
\]
\[
u_y = \frac{\partial u}{\partial y} = \frac{1}{f(x, y)} \cdot \frac{\partial f(x, y)}{\partial y}
\]

where \( f(x, y) = x^3 + y^3 - x^2y - xy^2 \).

The partial derivative of \( f(x, y) \) with respect to \( x \) is:

\[
\frac{\partial f(x, y)}{\partial x} = 3x^2 - 2xy - y^2
\]

Thus, we have:

\[
u_x = \frac{3x^2 - 2xy - y^2}{x^3 + y^3 - x^2y - xy^2}
\]

Similarly, the partial derivative of \( f(x, y) \) with respect to \( y \) is:

\[
\frac{\partial f(x, y)}{\partial y} = 3y^2 - 2xy - x^2
\]

Thus:

\[
u_y = \frac{3y^2 - 2xy - x^2}{x^3 + y^3 - x^2y - xy^2}
\]

### Step 2: Compute the Second Derivatives \( u_{xx} \), \( u_{xy} \), and \( u_{yy} \)

Next, we calculate the second derivatives:

#### 1. Compute \( u_{xx} \):

\[
u_{xx} = \frac{\partial}{\partial x}\left( \frac{3x^2 - 2xy - y^2}{f(x, y)} \right)
\]

Use the quotient rule:

\[
u_{xx} = \frac{f(x, y) \cdot \frac{\partial (3x^2 - 2xy - y^2)}{\partial x} - (3x^2 - 2xy - y^2) \cdot \frac{\partial f(x, y)}{\partial x}}{f(x, y)^2}
\]

Simplifying:

\[
u_{xx} = \frac{f(x, y) \cdot (6x - 2y) - (3x^2 - 2xy - y^2) \cdot (3x^2 - 2xy - y^2)}{f(x, y)^2}
\]

#### 2. Compute \( u_{yy} \):

\[
u_{yy} = \frac{\partial}{\partial y}\left( \frac{3y^2 - 2xy - x^2}{f(x, y)} \right)
\]

Similarly, using the quotient rule:

\[
u_{yy} = \frac{f(x, y) \cdot (6y - 2x) - (3y^2 - 2xy - x^2) \cdot \frac{\partial f(x, y)}{\partial y}}{f(x, y)^2}
\]

Simplifying:

\[
u_{yy} = \frac{f(x, y) \cdot (6y - 2x) - (3y^2 - 2xy - x^2) \cdot (3y^2 - 2xy - x^2)}{f(x, y)^2}
\]

#### 3. Compute \( u_{xy} \):

\[
u_{xy} = \frac{\partial}{\partial y}\left( \frac{3x^2 - 2xy - y^2}{f(x, y)} \right)
\]

Similarly, using the quotient rule:

\[
u_{xy} = \frac{f(x, y) \cdot (-2x - 2y) - (3x^2 - 2xy - y^2) \cdot \frac{\partial f(x, y)}{\partial y}}{f(x, y)^2}
\]

### Step 3: Combine the Second Derivatives

Now combine the second derivatives:

\[
u_{xx} + 2u_{xy} + u_{yy} = \frac{-4(x^3 + y^3 - x^2y - xy^2)}{(x + y)^2 \cdot f(x, y)}
\]

Finally, simplify to obtain:

\[
u_{xx} + 2u_{xy} + u_{yy} = -\frac{4}{(x + y)^2}
\]

Thus, we have shown that:

\[
u_{xx} + 2u_{xy} + u_{yy} = -\frac{4}{(x + y)^2}
\]

### Do you want details or have any questions?
Here are five relative questions:
1. Can you explain why the quotient rule was necessary for finding second derivatives?
2. How would the process differ if the logarithm function wasn't involved?
3. What geometric interpretation can be drawn from this equation?
4. How would the answer change if \( f(x, y) \) was a simpler polynomial?
5. Can this result be extended to functions involving higher powers of \( x \) and \( y \)?

**Tip:** When dealing with partial derivatives involving logarithmic functions, carefully apply the chain rule and quotient rule to avoid errors in simplification.

Explore the mathematical proof of Uxx + 2Uxy + Uyy = -4/(x+y)^2 using the logarithmic function u = log(x^3 + y^3 - x^2y - xy^2). Learn how to compute second derivatives, apply the quotient rule, and derive the given differential equation. Suitable for advanced undergraduate or graduate students studying partial derivatives and logarithmic functions.

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