1. **State the problem:** Find the derivative with respect to $x$ of the function $$f(x) = \frac{\ln x}{x^2}.$$\n\n2. **Formula used:** We use the quotient rule for derivatives, which states:\n$$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$\nwhere $u = \ln x$ and $v = x^2$.\n\n3. **Calculate derivatives of numerator and denominator:**\n$$\frac{du}{dx} = \frac{1}{x}$$\n$$\frac{dv}{dx} = 2x$$\n\n4. **Apply the quotient rule:**\n$$\frac{d}{dx} \left( \frac{\ln x}{x^2} \right) = \frac{x^2 \cdot \frac{1}{x} - \ln x \cdot 2x}{(x^2)^2} = \frac{x - 2x \ln x}{x^4}$$\n\n5. **Simplify the numerator:**\n$$x - 2x \ln x = x(1 - 2 \ln x)$$\n\n6. **Simplify the entire expression:**\n$$\frac{x(1 - 2 \ln x)}{x^4} = \frac{\cancel{x}(1 - 2 \ln x)}{\cancel{x} x^3} = \frac{1 - 2 \ln x}{x^3}$$\n\n**Final answer:**\n$$\boxed{\frac{d}{dx} \left( \frac{\ln x}{x^2} \right) = \frac{1 - 2 \ln x}{x^3}}$$