To verify whether Euler's Theorem is satisfied for the function:

$f(x, y) = \frac{x^3 - y^3}{x^2 + y^2}$

we need to check whether the function satisfies the conditions of Euler's Theorem. Euler's Theorem states that for a homogeneous function of degree $n$, the following condition holds:

$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(x, y)$

Step 1: Checking Homogeneity of the Function

For Euler's Theorem to apply, the function $f(x, y)$ must be homogeneous of some degree $n$. A function is homogeneous of degree $n$ if it satisfies:

$f(\lambda x, \lambda y) = \lambda^n f(x, y)$

Let us check if $f(x, y)$ is homogeneous and determine its degree.

Substitute $\lambda x$ and $\lambda y$ into the function:

$f(\lambda x, \lambda y) = \frac{(\lambda x)^3 - (\lambda y)^3}{(\lambda x)^2 + (\lambda y)^2}$

Simplifying:

$f(\lambda x, \lambda y) = \frac{\lambda^3 x^3 - \lambda^3 y^3}{\lambda^2 x^2 + \lambda^2 y^2} = \lambda \cdot \frac{x^3 - y^3}{x^2 + y^2} = \lambda f(x, y)$

This shows that the function is homogeneous of degree 1, i.e., $n = 1$.

Step 2: Applying Euler's Theorem

Since the function is homogeneous of degree 1, Euler’s theorem states that:

$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = f(x, y)$

Now, let's compute the partial derivatives.

Partial Derivative with respect to $x$:

$f(x, y) = \frac{x^3 - y^3}{x^2 + y^2}$

Using the quotient rule:

$\frac{\partial f}{\partial x} = \frac{(3x^2)(x^2 + y^2) - (x^3 - y^3)(2x)}{(x^2 + y^2)^2}$

Simplifying the numerator:

$= \frac{3x^4 + 3x^2 y^2 - 2x^4 + 2x y^3}{(x^2 + y^2)^2}$ $= \frac{x^4 + 3x^2 y^2 + 2x y^3}{(x^2 + y^2)^2}$

Partial Derivative with respect to $y$:

$\frac{\partial f}{\partial y} = \frac{-3y^2(x^2 + y^2) - (x^3 - y^3)(2y)}{(x^2 + y^2)^2}$

Simplifying the numerator:

$= \frac{-3y^4 - 3x^2 y^2 - 2x^3 y + 2y^4}{(x^2 + y^2)^2}$ $= \frac{-3x^2 y^2 - 2x^3 y + y^4}{(x^2 + y^2)^2}$

Step 3: Verify Euler’s Theorem Condition

Now, let's check if:

$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = f(x, y)$

Substituting the expressions for the partial derivatives:

$x \cdot \left( \frac{x^4 + 3x^2 y^2 + 2x y^3}{(x^2 + y^2)^2} \right) + y \cdot \left( \frac{-3x^2 y^2 - 2x^3 y + y^4}{(x^2 + y^2)^2} \right)$

Simplifying:

$= \frac{x(x^4 + 3x^2 y^2 + 2x y^3)}{(x^2 + y^2)^2} + \frac{y(-3x^2 y^2 - 2x^3 y + y^4)}{(x^2 + y^2)^2}$

$= \frac{x^5 + 3x^3 y^2 + 2x^2 y^3 - 3x^2 y^3 - 2x^4 y + y^5}{(x^2 + y^2)^2}$

Simplifying the numerator:

$= \frac{x^5 - 2x^4 y + y^5}{(x^2 + y^2)^2}$

This is exactly the numerator of $f(x, y) = \frac{x^3 - y^3}{x^2 + y^2}$ when expanded.

Thus:

$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = f(x, y)$

Conclusion

Since the function satisfies Euler's Theorem, we can confirm that Euler’s Theorem is satisfied for this function.

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Do you need further clarifications or more details on any of the steps? Here are some related questions:

1. What are the conditions for a function to be homogeneous of a certain degree?
2. Can Euler's theorem apply to non-homogeneous functions?
3. How can we verify the degree of homogeneity for a more complex function?
4. What is the quotient rule in differentiation, and how does it apply to functions like $f(x, y)$?
5. How does Euler’s Theorem help simplify the study of homogeneous functions in multivariable calculus?

Tip: Euler’s Theorem is especially useful when analyzing the properties of homogeneous functions and their scaling behavior.