1. Problem: Calculate the improper integral $$\int_{-1}^0 \frac{e^{1/x}}{x^3} \, dx$$. 2. We recognize the integral is improper at $x=0$ because the integrand involves $x^3$ in the denominator and $e^{1/x}$ which is undefined at 0. 3. Use substitution: let $$t = \frac{1}{x} \Rightarrow x = \frac{1}{t}$$. 4. Then $$dx = -\frac{1}{t^2} dt$$. 5. Change limits: when $x = -1$, $t = -1$; when $x \to 0^-$, $t \to -\infty$. 6. Substitute into the integral: $$\int_{-1}^0 \frac{e^{1/x}}{x^3} dx = \int_{t=-1}^{t \to -\infty} \frac{e^t}{(1/t)^3} \left(-\frac{1}{t^2}\right) dt = \int_{-1}^{-\infty} e^t \frac{t^3}{1} \left(-\frac{1}{t^2}\right) dt = \int_{-1}^{-\infty} -t e^t dt$$ 7. Reverse limits to get rid of negative sign: $$= \int_{-\infty}^{-1} t e^t dt$$ 8. Integrate by parts: Let $u = t$, $dv = e^t dt$; then $du = dt$, $v = e^t$. 9. So, $$\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t + C = e^t (t - 1) + C$$ 10. Evaluate the definite integral: $$\int_{-\infty}^{-1} t e^t dt = \lim_{a \to -\infty} [e^t (t - 1)]_a^{-1} = e^{-1}(-1 - 1) - \lim_{a \to -\infty} e^a (a - 1)$$ 11. As $a \to -\infty$, $e^a \to 0$ and $a - 1 \to -\infty$, but $e^a$ decays faster, so the limit is 0. 12. Thus, $$= e^{-1}(-2) - 0 = -\frac{2}{e}$$ Final answer: $$\boxed{-\frac{2}{e}}$$