1. The problem is to solve the differential equation $$\frac{dy}{dx} = 3x^{-2} e^{-y}$$ for $y$ as a function of $x$. 2. Separate variables to isolate $y$ terms on one side and $x$ terms on the other: $$e^y dy = 3x^{-2} dx$$ 3. Integrate both sides: $$\int e^y dy = \int 3x^{-2} dx$$ 4. The integral of $e^y$ with respect to $y$ is $e^y$, and the integral of $3x^{-2}$ with respect to $x$ is: $$3 \int x^{-2} dx = 3(-x^{-1}) + C = -\frac{3}{x} + C$$ 5. So we have: $$e^y = -\frac{3}{x} + C$$ 6. Take the natural logarithm of both sides to solve for $y$: $$y = \ln\left(-\frac{3}{x} + C\right)$$ 7. This is the implicit general solution to the differential equation, where $C$ is an arbitrary constant. Final answer: $$y = \ln\left(-\frac{3}{x} + C\right)$$