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The problem provided is:

Given \( u = \tan^{-1} \left( \frac{x^3 + y^3}{x + y} \right) \), find the expression \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \).

### Step-by-step solution:

1. **Let \( u = \tan^{-1}(z) \), where \( z = \frac{x^3 + y^3}{x + y} \)**. We need to compute the partial derivatives of \( u \) with respect to \( x \) and \( y \).

2. **Chain Rule for Derivatives**:
   \[
   \frac{\partial u}{\partial x} = \frac{1}{1+z^2} \cdot \frac{\partial z}{\partial x}, \quad \frac{\partial u}{\partial y} = \frac{1}{1+z^2} \cdot \frac{\partial z}{\partial y}
   \]

3. **Finding \( \frac{\partial z}{\partial x} \)**:
   \[
   z = \frac{x^3 + y^3}{x + y}
   \]
   Using the quotient rule:
   \[
   \frac{\partial z}{\partial x} = \frac{(3x^2)(x + y) - (x^3 + y^3)(1)}{(x + y)^2}
   \]
   Simplifying the numerator:
   \[
   = \frac{3x^2(x + y) - (x^3 + y^3)}{(x + y)^2}
   \]

4. **Finding \( \frac{\partial z}{\partial y} \)** similarly:
   \[
   \frac{\partial z}{\partial y} = \frac{(3y^2)(x + y) - (x^3 + y^3)(1)}{(x + y)^2}
   \]
   Simplifying the numerator:
   \[
   = \frac{3y^2(x + y) - (x^3 + y^3)}{(x + y)^2}
   \]

5. **Substitute into the expressions for \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \)**.

6. Finally, multiply the partial derivatives by \( x \) and \( y \) respectively, and sum:
   \[
   x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}
   \]

   After computation, this expression simplifies to:

   \[
   x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0
   \]

So, the result is \( \boxed{0} \).

Let me know if you'd like further explanation or clarifications!

### Follow-up Questions:
1. How does the quotient rule work in partial differentiation?
2. What is the geometric interpretation of the result \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0 \)?
3. Can the same approach be applied to similar trigonometric inverse functions?
4. What role does the chain rule play in multivariable calculus?
5. How does the tangent inverse function affect the overall derivative?

### Tip:
When differentiating inverse trigonometric functions, remember that the chain rule and quotient rule are key tools for handling complex expressions efficiently.

Given u = tan^(-1)((x^3 + y^3) / (x + y)), find x(∂u/∂x) + y(∂u/∂y).

Learn how to compute the partial derivatives and solve for the expression x(∂u/∂x) + y(∂u/∂y), given u = tan^(-1)((x^3 + y^3) / (x + y)). This solution involves applying the chain rule and quotient rule in multivariable calculus. Suitable for undergraduate calculus students.

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