We need to find the domain of the function $f(x) = \frac{x^2}{3 - 4x^4}$. The domain of a rational function is all real numbers except for the values of $x$ that make the denominator equal to zero.

1. Problem Description

We need to find the domain of the function $f(x) = \frac{x^2}{3 - 4x^4}$. The domain of a rational function is all real numbers except for the values of $x$ that make the denominator equal to zero.

2. Solution Steps

We need to find the values of $x$ for which the denominator $3 - 4x^4$ is equal to zero.

So, we set $3 - 4x^4 = 0$ and solve for $x$.

$3 - 4x^4 = 0$

$4x^4 = 3$

$x^4 = \frac{3}{4}$

$x = \pm \sqrt[4]{\frac{3}{4}}$

$x = \pm \frac{\sqrt[4]{3}}{\sqrt[4]{4}} = \pm \frac{\sqrt[4]{3}}{\sqrt{2}}$

$x = \pm \frac{\sqrt[4]{3}\sqrt{2}}{\sqrt{2}\sqrt{2}} = \pm \frac{\sqrt[4]{3}\sqrt{2}}{2}$

$x = \pm \frac{\sqrt{2\sqrt{3}}}{2}$

Therefore, the domain of the function is all real numbers except $x = \frac{\sqrt{2\sqrt{3}}}{2}$ and $x = -\frac{\sqrt{2\sqrt{3}}}{2}$.

In interval notation, the domain is $(-\infty, -\frac{\sqrt{2\sqrt{3}}}{2}) \cup (-\frac{\sqrt{2\sqrt{3}}}{2}, \frac{\sqrt{2\sqrt{3}}}{2}) \cup (\frac{\sqrt{2\sqrt{3}}}{2}, \infty)$.

3. Final Answer

The domain of the function is $(-\infty, -\frac{\sqrt{2\sqrt{3}}}{2}) \cup (-\frac{\sqrt{2\sqrt{3}}}{2}, \frac{\sqrt{2\sqrt{3}}}{2}) \cup (\frac{\sqrt{2\sqrt{3}}}{2}, \infty)$.