Examine the continuity of the following function 𝑓 at every point of its domain:
𝑓 (𝑥, 𝑦) =






𝑥
2
−
𝑦
3
𝑥
2
+𝑦
3
,
(𝑥, 𝑦) ≠ (0, 0)
0,
(𝑥, 𝑦) = (0, 0)

To examine the continuity of the function \( f(x, y) \) at every point in its domain, let's start by analyzing the function at the point \((0, 0)\) and then at other points in the domain.

### Function Definition
The function \( f(x, y) \) is defined as:
\[
f(x, y) =
\begin{cases} 
\frac{x^2 - y^3}{x^2 + y^3}, & \text{if } (x, y) \neq (0, 0) \\
0, & \text{if } (x, y) = (0, 0)
\end{cases}
\]

### Continuity at \((0, 0)\)
A function is continuous at a point if the limit as we approach that point is equal to the function's value at that point.

So, we need to check:
\[
\lim_{(x, y) \to (0, 0)} f(x, y)
\]
and see if it equals \( f(0, 0) \).

#### Evaluating the Limit
We'll evaluate the limit along different paths towards \((0, 0)\):

1. **Path \( y = 0 \)**:
   \[
   f(x, 0) = \frac{x^2 - 0^3}{x^2 + 0^3} = \frac{x^2}{x^2} = 1
   \]
   \[
   \lim_{x \to 0} f(x, 0) = 1
   \]

2. **Path \( x = 0 \)**:
   \[
   f(0, y) = \frac{0^2 - y^3}{0^2 + y^3} = \frac{-y^3}{y^3} = -1
   \]
   \[
   \lim_{y \to 0} f(0, y) = -1
   \]

3. **Path \( y = x^2 \)**:
   \[
   f(x, x^2) = \frac{x^2 - (x^2)^3}{x^2 + (x^2)^3} = \frac{x^2 - x^6}{x^2 + x^6} = \frac{x^2(1 - x^4)}{x^2(1 + x^4)} = \frac{1 - x^4}{1 + x^4}
   \]
   \[
   \lim_{x \to 0} f(x, x^2) = \frac{1 - 0}{1 + 0} = 1
   \]

These different paths give different limits (\(1\) along \(y = 0\) and \(y = x^2\), and \(-1\) along \(x = 0\)). Therefore, the limit as \((x, y) \to (0, 0)\) depends on the path taken, and hence **the limit does not exist**.

Since the limit does not exist, the function is **not continuous** at \((0, 0)\).

### Continuity at Points \((x, y) \neq (0, 0)\)
For any point \((x, y) \neq (0, 0)\), the function is defined as:
\[
f(x, y) = \frac{x^2 - y^3}{x^2 + y^3}
\]
Since this is a ratio of polynomials, and the denominator does not equal zero when \((x, y) \neq (0, 0)\), the function is continuous at all such points.

### Conclusion
- \( f(x, y) \) is **not continuous** at \((0, 0)\).
- \( f(x, y) \) is **continuous** at every other point in its domain.

Would you like further details or explanations on this analysis?

Here are 5 related questions you might find interesting:
1. How do you determine the continuity of a function at a given point using limits?
2. What are some common paths to consider when checking the limit of a multivariable function?
3. How does the behavior of a function differ at a point of discontinuity compared to a point of continuity?
4. Can you analyze the continuity of a piecewise function with more complex conditions?
5. What are the implications of discontinuity in practical applications, such as physics or engineering?

**Tip:** When evaluating limits in multivariable calculus, always check the limit along different paths to ensure the limit is path-independent. If the limit varies based on the path, it does not exist.

Explore the continuity of the multivariable function f(x, y) = (x^2 - y^3) / (x^2 + y^3) at every point in its domain. Detailed analysis includes limit evaluations along different paths towards (0, 0). Suitable for college-level students studying multivariable calculus.

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