We need to find the limit of the function $f(x, y) = \frac{xy^2}{x^2 + y^4}$ as $(x, y)$ approaches $(0, 0)$.

1. Problem Description

We need to find the limit of the function $f(x, y) = \frac{xy^2}{x^2 + y^4}$ as $(x, y)$ approaches $(0, 0)$.

2. Solution Steps

To determine if the limit exists, we will analyze the limit along different paths approaching $(0, 0)$.

Path 1: $y = mx$

Substitute $y = mx$ into the expression:

$\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2 + y^4} = \lim_{x \to 0} \frac{x(mx)^2}{x^2 + (mx)^4} = \lim_{x \to 0} \frac{m^2x^3}{x^2 + m^4x^4} = \lim_{x \to 0} \frac{m^2x}{1 + m^4x^2} = \frac{m^2(0)}{1 + m^4(0)^2} = \frac{0}{1} = 0$.

Path 2: $x = y^2$

Substitute $x = y^2$ into the expression:

$\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2 + y^4} = \lim_{y \to 0} \frac{y^2y^2}{(y^2)^2 + y^4} = \lim_{y \to 0} \frac{y^4}{y^4 + y^4} = \lim_{y \to 0} \frac{y^4}{2y^4} = \lim_{y \to 0} \frac{1}{2} = \frac{1}{2}$.

Since the limit along the path $y = mx$ is 0, and the limit along the path $x = y^2$ is $\frac{1}{2}$, the limit does not exist.

3. Final Answer

The limit does not exist.