1. **State the problem:** Find the derivative $f'(x)$ of the function $$f(x) = \frac{x^3}{\tan x}.$$\n\n2. **Recall the formula:** To differentiate a quotient $\frac{u}{v}$, use the quotient rule: $$f'(x) = \frac{u'v - uv'}{v^2}$$ where $u = x^3$ and $v = \tan x$.\n\n3. **Find derivatives of numerator and denominator:**\n- $u' = \frac{d}{dx} x^3 = 3x^2$\n- $v' = \frac{d}{dx} \tan x = \sec^2 x$\n\n4. **Apply the quotient rule:**\n$$f'(x) = \frac{3x^2 \cdot \tan x - x^3 \cdot \sec^2 x}{(\tan x)^2}.$$\n\n5. **Simplify the expression:**\nThe derivative is $$f'(x) = \frac{3x^2 \tan x - x^3 \sec^2 x}{\tan^2 x}.$$\n\nThis is the simplified form of the derivative.\n\n**Final answer:** $$\boxed{f'(x) = \frac{3x^2 \tan x - x^3 \sec^2 x}{\tan^2 x}}.$$