I follow your work fine up through this step:
\(\displaystyle \lim_{x \to \infty} \frac{\ln \left( \frac{1}{1+ \frac{1}{x}} \right)}{\frac{1}{x}}\)
But after that you lose me completely. You write that you're applying L'Hopital's Rule, but that doesn't look like any application of the rule I've ever seen. You should end up with:
\(\displaystyle \lim_{x \to \infty} \frac{\frac{d}{dx} \left[ \ln \left( \frac{1}{1+ \frac{1}{x}} \right) \right]}{\frac{d}{dx} \left[ \frac{1}{x} \right]}\)
\(\displaystyle = \lim_{x \to \infty} \frac{\frac{1}{x^2+x}}{-\frac{1}{x^2}}\)
When you simply that down, you'll get:
\(\displaystyle \lim_{x \to \infty} \ln(y(x)) = -1 \implies \lim_{x \to \infty} y(x) = e^{-1} = \frac{1}{e}\)