We are solving the limit:

$\lim_{(x, y) \to (0, 0)} \frac{x^3 y}{x^9 + y^3}$

with the substitution $y = x^3$.

---

Step 1: Substitution

Replace $y$ with $x^3$ in the given expression:

$\frac{x^3 y}{x^9 + y^3} = \frac{x^3 (x^3)}{x^9 + (x^3)^3}$

$= \frac{x^3 x^3}{x^9 + x^9}.$

---

Step 2: Simplify

Simplify the numerator and the denominator:

$\frac{x^3 x^3}{x^9 + x^9} = \frac{x^6}{2x^9}.$

$= \frac{1}{2x^3}.$

---

Step 3: Take the Limit

As $x \to 0$, the term $\frac{1}{2x^3}$ becomes undefined because $x^3$ in the denominator approaches $0$, leading to infinity.

Conclusion

The limit does not exist (it diverges to infinity).

---

Do you want a detailed explanation of this result, or have further questions? Here are related explorations:

1. What happens if we try another path, like $y = kx$?
2. How can we determine if a limit exists at a multivariable point in general?
3. What is the geometric interpretation of divergence in this example?
4. Can we use polar coordinates to analyze the limit?
5. What does it mean for a multivariable function to be continuous or discontinuous?

Tip: In multivariable calculus, always test the limit along multiple paths to check for consistency!