L’Hospital’s rule applies only to limits of the form $0/0$ and $\infty/\infty$. However, there is a standard trick for converting the indeterminate form $1^\infty$ to one of these forms so that l’Hospital’s rule can be applied, and that’s essentially what you’re doing here. Let me write it up in a way that makes a little clearer exactly what is going on.
Let $L=\displaystyle\lim_{x\to 0}\left(e^x+x\right)^{1/x}$. Then $$\ln L=\ln\lim_{x\to 0}\left(e^x+x\right)^{1/x}=\lim_{x\to 0}\ln\left(e^x+x\right)^{1/x}\;,$$ since $\ln$ is a continuous function. Thus,
$$\ln L=\lim_{x\to 0}\ln\left(e^x+x\right)^{1/x}=\lim_{x\to 0}\frac1x\ln(e^x+x)=\lim_{x\to 0}\frac{\ln(e^x+x)}x\;.$$
This last limit is a $\frac00$ indeterminate form, so l’Hospital’s rule applies:
$$\ln L=\lim_{x\to 0}\frac{\ln(e^x+x)}x=\lim_{x\to 0}\frac{\frac{e^x+1}{e^x+x}}1=\lim_{x\to 0}\frac{e^x+1}{e^x+x}=\frac21=2\;.$$
Finally, then $L=e^{\ln L}=e^2$.
You went astray when you wrote $$\lim_{x \to 0} \frac{\ln\left(e^x+x\right)}{x} = \lim_{x \to 0} \frac{1}{x^2+xe^x}\;:$$ you did not differentiate the numerator correctly, and you did not differentiate the denominator at all.
If $\lim_{x\to a}f(x)=0$ and $\lim_{x\to a}g(x)=1$, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac01=0$; there is nothing indeterminate here.
Best Answer
You may proceed as follows:
You have
$$n^3\ln (1+\frac 1{n!})= \frac{n^3}{n!}\ln (1+\frac 1{n!})^{n!}\stackrel{n\to\infty}{\longrightarrow}0\cdot 1=0$$
Here, I use $\lim_{m\to \infty}(1+\frac 1m)^m = e$ and $\ln e = 1$.