Solution: Since $ \sin^2 x + \cos^2 x = 1 $, we write $ f(x) = \frac{1}{\sin^2 x + 4\cos^2 x} = \frac{1}{1 + 3\cos^2 x} $. Let $ y = \cos^2 x $, so $ 0 \leq y \leq 1 $. Then $ f(x) = \frac{1}{1 + 3y} $. As $ y $ varies from 0 to 1, $ 1 + 3y $ varies from 1 to 4, so $ f(x) $ varies from $ \frac{1}{4} $ to $ 1 $. Hence, the range of $ f(x) $ is $ \left[\frac{1}{4}, 1\right] $. The final answer is $ \boxed{\left[\frac{1}{4}, 1\right]} $.

["Understanding and Finding the Range of a Trigonometric Function: Detailed Solution", "Since we know the fundamental identity $ \sin^2 x + \cos^2 x = 1 $, we can simplify complex trigonometric expressions and determine their ranges effectively. Consider the function:\n$$\nf(x) = \frac{1}{\sin^2 x + 4\cos^2 x}\n$$", "We begin by using the identity:\n$$\n\sin^2 x = 1 - \cos^2 x\n$$\nSubstituting this into the denominator:\n$$\n\sin^2 x + 4\cos^2 x = (1 - \cos^2 x) + 4\cos^2 x = 1 + 3\cos^2 x\n$$\nLet $ y = \cos^2 x $. Since $ \cos^2 x $ ranges between 0 and 1 for all real $ x $, we have $ 0 \leq y \leq 1 $. Therefore,\n$$\nf(x) = \frac{1}{1 + 3y}\n$$", "Now analyze $ f(x) $ over the interval $ y \in [0, 1] $:\n- When $ y = 0 $, $ f(x) = \frac{1}{1 + 0} = 1 $\n- When $ y = 1 $, $ f(x) = \frac{1}{1 + 3} = \frac{1}{4} $", "Because $ 1 + 3y $ increases steadily from 1 to 4 as $ y $ goes from 0 to 1, the function $ \frac{1}{1 + 3y} $ decreases continuously from 1 to $ \frac{1}{4} $. Thus, the output values of $ f(x) $ lie within the closed interval $ \left[\frac{1}{4}, 1\right] $.", "Therefore, the range of $ f(x) $ is\n$$\n\boxed{\left[\frac{1}{4}, 1\right]}\n$$", "This structured approach not only solves the problem clearly but also strengthens understanding of trigonometric identities and function behavior across domains."]