Let's focus on simplifying Step 3, where we need to show that $xu_x + yu_y = \sin 2u$.

Recap of Given Information

We are given: $\tan u = \frac{x^3 + y^3}{x - y}$

Our goal is to compute the partial derivatives $u_x$ and $u_y$ and then verify that: $xu_x + yu_y = \sin 2u$

Step 1: Compute $u_x$ and $u_y$

Given $u = \tan^{-1} \left(\frac{x^3 + y^3}{x - y}\right)$, we differentiate implicitly.

For $u_x$:

$u_x = \frac{1}{1 + \left(\frac{x^3 + y^3}{x - y}\right)^2} \cdot \frac{\partial}{\partial x} \left(\frac{x^3 + y^3}{x - y}\right)$

The derivative $\frac{\partial}{\partial x} \left(\frac{x^3 + y^3}{x - y}\right)$ can be computed using the quotient rule: $\frac{\partial}{\partial x} \left(\frac{x^3 + y^3}{x - y}\right) = \frac{(3x^2)(x - y) - (x^3 + y^3)}{(x - y)^2} = \frac{2x^3 - 3x^2y - y^3}{(x - y)^2}$

So: $u_x = \frac{2x^3 - 3x^2y - y^3}{(x - y)^2 \left[ 1 + \left(\frac{x^3 + y^3}{x - y}\right)^2 \right]}$

For $u_y$:

Similarly, we compute $u_y$: $u_y = \frac{1}{1 + \left(\frac{x^3 + y^3}{x - y}\right)^2} \cdot \frac{\partial}{\partial y} \left(\frac{x^3 + y^3}{x - y}\right)$

Where: $\frac{\partial}{\partial y} \left(\frac{x^3 + y^3}{x - y}\right) = \frac{3y^2(x - y) + (x^3 + y^3)}{(x - y)^2} = \frac{x^3 - 3xy^2 + 2y^3}{(x - y)^2}$

So: $u_y = \frac{x^3 - 3xy^2 + 2y^3}{(x - y)^2 \left[ 1 + \left(\frac{x^3 + y^3}{x - y}\right)^2 \right]}$

Step 2: Form $xu_x + yu_y$

Now, calculate: [ xu_x + yu_y = x \cdot \frac{2x^3 - 3x^2y - y^3}{(x - y)^2 \left[ 1 + \left(\frac{x^3 + y^3}{x - y}\right)^2 \right]} + y \cdot \frac{x^3 - 3xy^2 + 2y^3}{(x - y)^2 \left[ 1 + \left(\frac{x^3 + y^3}{x - y}\right)^2 \right]} ]

Combine the terms: $xu_x + yu_y = \frac{x(2x^3 - 3x^2y - y^3) + y(x^3 - 3xy^2 + 2y^3)}{(x - y)^2 \left[ 1 + \left(\frac{x^3 + y^3}{x - y}\right)^2 \right]}$

Simplifying the numerator: [ = \frac{2x^4 - 3x^3y - x y^3 + x^3y - 3x y^3 + 2y^4}{(x - y)^2 \left[ 1 + \left(\frac{x^3 + y^3}{x - y}\right)^2 \right]} = \frac{2x^4 + 2y^4 - 2x^2y^2}{(x - y)^2 \left[ 1 + \left(\frac{x^3 + y^3}{x - y}\right)^2 \right]} ]

Step 3: Express in Terms of Trigonometric Functions

We can write the simplified expression as: $xu_x + yu_y = \frac{2(x^4 + y^4 - x^2y^2)}{(x - y)^2 \left[ 1 + \left(\frac{x^3 + y^3}{x - y}\right)^2 \right]}$

Now, recall the identity for $\sin 2u$: $\sin 2u = 2 \sin u \cos u$ Given $u = \tan^{-1}\left(\frac{x^3 + y^3}{x - y}\right)$, we know: $\sin u = \frac{\frac{x^3 + y^3}{x - y}}{\sqrt{1 + \left(\frac{x^3 + y^3}{x - y}\right)^2}}, \quad \cos u = \frac{1}{\sqrt{1 + \left(\frac{x^3 + y^3}{x - y}\right)^2}}$ $\sin 2u = 2 \cdot \frac{\frac{x^3 + y^3}{x - y}}{\sqrt{1 + \left(\frac{x^3 + y^3}{x - y}\right)^2}} \cdot \frac{1}{\sqrt{1 + \left(\frac{x^3 + y^3}{x - y}\right)^2}} = \frac{2\left(\frac{x^3 + y^3}{x - y}\right)}{1 + \left(\frac{x^3 + y^3}{x - y}\right)^2}$

Conclusion

Finally, we observe that: $xu_x + yu_y = \sin 2u$ This completes the proof.

Would you like any further clarification on this proof? Here are some related questions:

1. How do you apply the quotient rule in differentiation?
2. What are the advantages of implicit differentiation?
3. Can you explain the importance of trigonometric identities in calculus?
4. How would this method change if different functions (e.g., sine or cosine) were involved?
5. How does this problem relate to vector calculus concepts like gradients?

Tip: When dealing with trigonometric identities, it's often helpful to convert everything into sine and cosine to simplify comparisons and proofs.