Preliminary Results:
We will use
$$
\begin{align}
\frac{\color{#C00000}{\sin(2x)-2\sin(x)}}{\color{#00A000}{\tan(2x)-2\tan(x)}}
&=\underbrace{\color{#C00000}{2\sin(x)(\cos(x)-1)}\vphantom{\frac{\tan^2(x)}{\tan^2(x)}}}\underbrace{\frac{\color{#00A000}{1-\tan^2(x)}}{\color{#00A000}{2\tan^3(x)}}}\\
&=\hphantom{\sin}\frac{-2\sin^3(x)}{\cos(x)+1}\hphantom{\sin}\frac{\cos(x)\cos(2x)}{2\sin^3(x)}\\
&=-\frac{\cos(x)\cos(2x)}{\cos(x)+1}\tag{1}
\end{align}
$$
Therefore,
$$
\lim_{x\to0}\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}=-\frac12\tag{2}
$$
Thus, given an $\epsilon\gt0$, we can find a $\delta\gt0$ so that if $|x|\le\delta$
$$
\left|\,\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}+\frac12\,\right|\le\epsilon\tag{3}
$$
Because $\,\displaystyle\lim_{x\to0}\frac{\sin(x)}{x}=\lim_{x\to0}\frac{\tan(x)}{x}=1$, we have
$$
\sin(x)-x=\sum_{k=0}^\infty2^k\sin(x/2^k)-2^{k+1}\sin(x/2^{k+1})\tag{4}
$$
and
$$
\tan(x)-x=\sum_{k=0}^\infty2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1})\tag{5}
$$
By $(3)$ each term of $(4)$ is between $-\frac12-\epsilon$ and $-\frac12+\epsilon$ of the corresponding term of $(5)$. Therefore,
$$
\left|\,\frac{\sin(x)-x}{\tan(x)-x}+\frac12\,\right|\le\epsilon\tag{6}
$$
Thus,
$$
\lim_{x\to0}\,\frac{\sin(x)-x}{\tan(x)-x}=-\frac12\tag{7}
$$
Furthermore,
$$
\begin{align}
\frac{\tan(x)-\sin(x)}{x^3}
&=\tan(x)(1-\cos(x))\frac1{x^3}\\
&=\frac{\sin(x)}{\cos(x)}\frac{\sin^2(x)}{1+\cos(x)}\frac1{x^3}\\
&=\frac1{\cos(x)(1+\cos(x))}\left(\frac{\sin(x)}{x}\right)^3\tag{8}
\end{align}
$$
Therefore,
$$
\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}=\frac12\tag{9}
$$
Combining $(7)$ and $(9)$ yield
$$
\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag{10}
$$
Additionally,
$$
\frac{\sin(A)-\sin(B)}{\sin(A-B)}
=\frac{\cos\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}
=1-\frac{2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}\tag{11}
$$
Finishing Up:
$$
\begin{align}
&x\sin(\sin(x))-\sin^2(x)\\
&=[\color{#C00000}{(x-\sin(x))+\sin(x)}][\color{#00A000}{(\sin(\sin(x))-\sin(x))+\sin(x)}]-\sin^2(x)\\
&=\color{#C00000}{(x-\sin(x))}\color{#00A000}{(\sin(\sin(x))-\sin(x))}\\
&+\color{#C00000}{(x-\sin(x))}\color{#00A000}{\sin(x)}\\
&+\color{#C00000}{\sin(x)}\color{#00A000}{(\sin(\sin(x))-\sin(x))}\\
&=(x-\sin(x))(\sin(\sin(x))-\sin(x))+\sin(x)(x-2\sin(x)+\sin(\sin(x)))\tag{12}
\end{align}
$$
Using $(10)$, we get that
$$
\begin{align}
&\lim_{x\to0}\frac{(x-\sin(x))(\sin(\sin(x))-\sin(x))}{x^6}\\
&=\lim_{x\to0}\frac{x-\sin(x)}{x^3}\lim_{x\to0}\frac{\sin(\sin(x))-\sin(x)}{\sin^3(x)}\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^3\\
&=\frac16\cdot\frac{-1}6\cdot1\\
&=-\frac1{36}\tag{13}
\end{align}
$$
and with $(10)$ and $(11)$, we have
$$
\begin{align}
&\lim_{x\to0}\frac{\sin(x)(x-2\sin(x)+\sin(\sin(x)))}{x^6}\\
&=\lim_{x\to0}\frac{\sin(x)}{x}\lim_{x\to0}\frac{x-2\sin(x)+\sin(\sin(x))}{x^5}\\
&=\lim_{x\to0}\frac{(x-\sin(x))-(\sin(x)-\sin(\sin(x))}{x^5}\\
&=\lim_{x\to0}\frac{(x-\sin(x))-\sin(x-\sin(x))\left(1-\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{\cos\left(\frac{x-\sin(x)}{2}\right)}\right)}{x^5}\\
&=\lim_{x\to0}\frac{(x-\sin(x))-\sin(x-\sin(x))+\sin(x-\sin(x))\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{\cos\left(\frac{x-\sin(x)}{2}\right)}}{x^5}\\
&=\lim_{x\to0}\frac{\sin(x-\sin(x))}{x^3}\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{x^2}\\[6pt]
&=\frac16\cdot\frac12\\[6pt]
&=\frac1{12}\tag{14}
\end{align}
$$
Adding $(13)$ and $(14)$ gives
$$
\color{#C00000}{\lim_{x\to0}\frac{x\sin(\sin(x))-\sin^2(x)}{x^6}=\frac1{18}}\tag{15}
$$
Added Explanation for the Derivation of $(6)$
The explanation below works for $x\gt0$ and $x\lt0$. Just reverse the red inequalities.
Assume that $x\color{#C00000}{\gt}0$ and $|x|\lt\pi/2$. Then $\tan(x)-2\tan(x/2)\color{#C00000}{\gt}0$.
$(3)$ is equivalent to
$$
\begin{align}
&(-1/2-\epsilon)(\tan(x)-2\tan(x/2))\\[4pt]
\color{#C00000}{\le}&\sin(x)-2\sin(x/2)\\[4pt]
\color{#C00000}{\le}&(-1/2+\epsilon)(\tan(x)-2\tan(x/2))\tag{16}
\end{align}
$$
for all $|x|\lt\delta$. Thus, for $k\ge0$,
$$
\begin{align}
&(-1/2-\epsilon)(2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1}))\\[4pt]
\color{#C00000}{\le}&2^k\sin(x/2^k)-2^{k+1}\sin(x/2^{k+1})\\[4pt]
\color{#C00000}{\le}&(-1/2+\epsilon)(2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1}))\tag{17}
\end{align}
$$
Summing $(17)$ from $k=0$ to $\infty$ yields
$$
\begin{align}
&(-1/2-\epsilon)\left(\tan(x)-\lim_{k\to\infty}2^k\tan(x/2^k)\right)\\[4pt]
\color{#C00000}{\le}&\sin(x)-\lim_{k\to\infty}2^k\sin(x/2^k)\\[4pt]
\color{#C00000}{\le}&(-1/2+\epsilon)\left(\tan(x)-\lim_{k\to\infty}2^k\tan(x/2^k)\right)\tag{18}
\end{align}
$$
Since $\lim\limits_{k\to\infty}2^k\tan(x/2^k)=\lim\limits_{k\to\infty}2^k\sin(x/2^k)=x$, $(18)$ says
$$
\begin{align}
&(-1/2-\epsilon)(\tan(x)-x)\\[4pt]
\color{#C00000}{\le}&\sin(x)-x\\[4pt]
\color{#C00000}{\le}&(-1/2+\epsilon)(\tan(x)-x))\tag{19}
\end{align}
$$
which, since $\epsilon$ is arbitrary is equivalent to $(6)$.
Notice that $\frac{\arcsin x}{x} \to 1$ as $x \to 0$, then
$$\ln L = \lim_{x \to 0} \frac{\ln\left(1+\frac{\arcsin x}{x}-1\right)}{x^2} = \lim_{x \to 0} \frac{\frac{\arcsin x}{x}-1}{x^2} = \lim_{x \to 0} \frac{\arcsin x-x}{x^3}.$$
Taking $x = \sin t$, the limit becomes
$$\ln L = \lim_{t \to 0} \frac{t - \sin t}{\sin^3 t} = \lim_{t \to 0} \frac{t - \sin t}{t^3} = \lim_{t \to 0} \frac{1 - \cos t}{3t^2} = \frac{1}{6},$$
and hence $L = e^{1/6}$.
Note: Such method is useful to most of $1^{\infty}$-form limits.
Best Answer
$$\begin{align*} & \lim_{x \rightarrow 0} \frac{e^x - e^{x\cos x}}{x + \sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x - 1 + 1 - e^{x\cos x}}{x + \sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x+\sin x} + \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{x+\sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \frac{x}{x+\sin x} + \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{-x\cos x} \cdot \frac{-x\cos x}{x+\sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \lim_{x \rightarrow 0} \frac{1}{1 + \frac{\sin x}{x}} + \lim_{x\cos x \rightarrow 0} \frac{e^{x\cos x}-1}{x\cos x} \cdot \lim_{x \rightarrow 0} -\frac{x}{x+\sin x} \cdot \cos x \\ & = 1\cdot \frac{1}{1+1} + 1 \cdot -\frac{1}{1+1} \cdot 1\\ & = \frac{1}{2} - \frac{1}{2}\\ & = 0 \\ \end{align*}$$