1. **Problem:** Find the derivative of the function $$f(x) = \frac{\exp(x)}{x^2}, x > 0.$$\n\n2. **Formula:** Use the quotient rule for derivatives: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}.$$ Here, $u = \exp(x)$ and $v = x^2$.\n\n3. **Calculate derivatives:**\n- $u' = \frac{d}{dx} \exp(x) = \exp(x)$\n- $v' = \frac{d}{dx} x^2 = 2x$\n\n4. **Apply quotient rule:**\n$$f'(x) = \frac{\exp(x) \cdot x^2 - \exp(x) \cdot 2x}{(x^2)^2} = \frac{\exp(x)(x^2 - 2x)}{x^4}.$$\n\n5. **Simplify:**\n$$f'(x) = \exp(x) \frac{x^2 - 2x}{x^4} = \exp(x) \frac{x(x - 2)}{x^4} = \exp(x) \frac{x - 2}{x^3}.$$\n\n**Final answer:** $$\boxed{f'(x) = \exp(x) \frac{x - 2}{x^3}}.$$