$$\lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}.$$
\begin{align*} &\lim_{x \to 0}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}\\ =&\lim_{x \to 0}\frac{\exp [(\sin x)^x\ln x]-\exp [{x^{\sin x}\ln(\sin x)]}}{x^3} \\=&\lim_{x \to 0}\frac{\exp [{x^{\sin x}\ln(\sin x)]}(\exp [(\sin x)^x\ln x-{x^{\sin x}\ln(\sin x)}]-1)}{x^3} \\=&\lim_{x \to 0}\frac{\exp [{x^{\sin x}\ln(\sin x)]} [(\sin x)^x\ln x-{x^{\sin x}\ln(\sin x)}]}{x^3} \end{align*} This will help?
Note that $$ \begin{align} &\log\left(\frac{x^{\sin(x)^x}}x\right)\\[3pt] &=\log(x)\sin(x)^x-\log(x)\\[6pt] &=\color{#C00}{\log(x)}\,\color{#090}{x^x}\color{#00F}{\left(\frac{\sin(x)}x\right)^x}-\log(x)\\ &=\color{#C00}{\log(x)}\color{#090}{\left(1+x\log(x)+\frac{x^2\log(x)^2}2+\varepsilon_{3,3}\right)}\color{#00F}{\left(1+\varepsilon_{3,0}\right)}-\log(x)\\ &=x\log(x)^2+\frac{x^2\log(x)^3}2+\varepsilon_{3,4} \end{align} $$ and $$ \begin{align} &\log\left(\frac{\sin(x)^{x^{\sin(x)}}}x\right)\\[3pt] &=\log(\sin(x))\,x^{\sin(x)}-\log(x)\\[9pt] &=\color{#C00}{\log(\sin(x))}\,\color{#090}{x^x}\color{#00F}{x^{\sin(x)-x}}-\log(x)\\[3pt] &=\color{#C00}{\left(\log(x)-\frac{x^2}6+\varepsilon_{4,0}\right)}\color{#090}{\left(1+x\log(x)+\frac{x^2\log(x)^2}2+\varepsilon_{3,3}\right)}\color{#00F}{\left(1+\varepsilon_{3,1}\right)}-\log(x)\\ &=x\log(x)^2+\frac{x^2\log(x)^3}2-\frac{x^2}6+\varepsilon_{3,4} \end{align} $$ where $\varepsilon_{m,n}=O\!\left(x^m\log(x)^n\right)$.
If $u,v\to0$, then $e^u-e^v=(u-v)\left(1+\frac{u+v}2\right)+O\!\left(u^3\right)+O\!\left(v^3\right)$. Thus, $$ \frac{x^{\sin(x)^x}}x-\frac{\sin(x)^{x^{\sin(x)}}}x=\frac{x^2}6+\varepsilon_{3,6} $$ and therefore, $$ \frac{x^{\sin(x)^x}-\sin(x)^{x^{\sin(x)}}}{x^3}=\frac16+\varepsilon_{1,6} $$ That is, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{x^{\sin(x)^x}-\sin(x)^{x^{\sin(x)}}}{x^3}=\frac16} $$
In General
Note that $$ \begin{align} \left(x+O\!\left(x^3\right)\right)^{x+O\left(x^3\right)} &=\color{#C00}{x^x}\color{#090}{x^{O\left(x^3\right)}}\color{#00F}{\left(1+O\!\left(x^2\right)\right)^{x+O\left(x^3\right)}}\\[3pt] &=\color{#C00}{\left(1+x\log(x)+\tfrac12x^2\log(x)^2+\varepsilon_{3,3}\right)}\color{#090}{(1+\varepsilon_{3,1})}\color{#00F}{(1+\varepsilon_{3,0})}\\[3pt] &=1+x\log(x)+\tfrac12x^2\log(x)^2+\varepsilon_{3,3}\tag1 \end{align} $$ where $\varepsilon_{m,n}=O\!\left(x^m\log(x)^n\right)$.
Furthermore, $$ \begin{align} \log\left(x+ax^3+o\!\left(x^3\right)\right) &=\log(x)+\log\left(1+ax^2+o\!\left(x^2\right)\right)\\[3pt] &=\log(x)+ax^2+o\!\left(x^2\right)\tag2 \end{align} $$ Therefore, $$ \begin{align} &\log\left(x+ax^3+o\!\left(x^3\right)\right)\left(x+O\!\left(x^3\right)\right)^{x+O\left(x^3\right)}-\log(x)\\[3pt] &=x\log(x)^2+\tfrac12x^2\log(x)^3+ax^2+o\!\left(x^2\right)\tag3 \end{align} $$ If $u,v\to0$, then $e^u-e^v=(u-v)\left(1+\frac{u+v}2\right)+O\!\left(u^3\right)+O\!\left(v^3\right)$.
Thus, $$ \begin{align} &\frac1x\left[\left(x+a_1x^3+o\!\left(x^3\right)\right)^{\left(x+O\left(x^3\right)\right)^{x+O\left(\lower{.5pt}{x^3}\right)}} -\left(x+a_2x^3+o\!\left(x^3\right)\right)^{\left(x+O\left(x^3\right)\right)^{x+O\left(\lower{.5pt}{x^3}\right)}}\right]\\[6pt] &=(a_1-a_2)x^2+o\!\left(x^2\right)\tag4 \end{align} $$ Therefore, $$ \bbox[5px,border:2px solid #C0A000]{\begin{align} &\lim_{x\to0}\frac1{x^3}\left[\left(x+a_1x^3+o\!\left(x^3\right)\right)^{\left(x+O\left(x^3\right)\right)^{x+O\left(\lower{.5pt}{x^3}\right)}} -\left(x+a_2x^3+o\!\left(x^3\right)\right)^{\left(x+O\left(x^3\right)\right)^{x+O\left(\lower{.5pt}{x^3}\right)}}\right]\\[6pt] &=a_1-a_2 \end{align}}\tag5 $$ and this verifies the other limits in the similar question (which wanted an answer without Taylor series).
Too long for a comment (I don't have the points so...)
We have :
$$\lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}=\lim_{x \to 0+}\frac{-x^{(\sin x)^x}+(Arcsin (x))^{(x)^{Arcsin (x)}}}{x^3}.$$
So we introduce the following function : $$f(x)=x^{sin(x)^x}$$
We have to prove :
$$\lim_{x \to 0+}\frac{-f(x)+f(Arcsin(x))}{x^3}$$
Or $$\lim_{x \to 0+}\frac{(-f(x)+f(Arcsin(x)))(Arcsin(x)-x)}{(Arcsin(x)-x)x^3}$$
But :
$$\lim_{x \to 0+}\frac{(Arcsin(x)-x)}{x^3}=\frac{1}{6}$$
And
$$\lim_{x \to 0+}\frac{(-f(x)+f(Arcsin(x)))}{(Arcsin(x)-x)}=1$$
For the last limit I use the three chord lemma and the fact that the function $f(x)$ is convex on $[0;\varepsilon]$ for $\varepsilon$ very small (near from 0).
Recall the following formulas $$\sin x=x-\frac{1}{3!}x^3+\frac{1}{5!}x-\cdots\tag 1$$ $$e^x=1+x+\frac{1}{2}x^2+\cdots\tag 2$$ $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots \tag 3$$ which are the starting point we need to rely on.
From$(1)$, \begin{align} \frac{\sin x-x}{x}=\dfrac{x-\dfrac{1}{3!}x^3+o(x^4)-x}{x}=-\frac{1}{6}x^2+o(x^3).\tag{4} \end{align} From $(3)(4)$, \begin{align} \ln\sin x&=\ln\left(x\cdot \frac{\sin x}{x}\right)=\ln x+\ln\left(1+\frac{\sin x-x}{x}\right)\notag\\ &=\ln x+\frac{\sin x-x}{x}-\frac{1}{2}\left(\frac{\sin x-x}{x}\right)^2+o(x^4)\notag\\ &=\ln x-\frac{1}{6}x^2+o(x^3).\tag{5} \end{align} From $(5)$, $$x\ln\sin x=x(\ln x-\frac{1}{6}x^2+o(x^3))=x\ln x-\frac{1}{6}x^3+o(x^3).\tag{6}$$ From $(2)(6)$, \begin{align*} (\sin x)^x&=\exp(x\ln\sin x)=1+x\ln\sin x+\frac{1}{2}(x\ln\sin x)^2+\frac{1}{6}(x\ln\sin x)^3+o(x^3\ln^3 x)\notag\\ &=1+x\ln x+\frac{1}{2}x^2\ln^2 x+\frac{1}{6}x^3\ln^3 x+o(x^3\ln^3 x).\tag{7} \end{align*} From $(7)$, we obtain a vital result that \begin{align*} {\color{red}{f(x):=(\sin x)^x\ln x}}=\ln x+x\ln^2 x+\frac{1}{2}x^2\ln^3 x+\frac{1}{6}x^3\ln^4 x+o(x^3\ln^4 x).\tag{8} \end{align*}
Morover, by $(1)(2)$, we have \begin{align*} x^{\sin x}&=\exp(\sin x\ln x)=1+\sin x\ln x+\frac{1}{2}\sin^2 x\ln^2 x+\frac{1}{6}\sin^3 x\ln^3 x+o(x^3\ln^3 x)\notag\\ &=1+x\ln x+\frac{1}{2} x^2\ln^2 x+\frac{1}{6}x^3\ln^3 x+o(x^3\ln^3 x).\tag{9} \end{align*} From $(5)(9)$, we oabtain another vital result that \begin{align*} &{\color{red}{g(x):=x^{\sin x}\ln\sin x}}\\ =&(1+x\ln x+\frac{1}{2} x^2\ln^2 x+\frac{1}{6}x^3\ln^3 x+o(x^3\ln^3 x))(\ln x-\frac{1}{6}x^2+o(x^3))\\ =&\ln x+x\ln^2 x+\frac{1}{2}x^2\ln^3 x+\frac{1}{6}x^3\ln^4 x-\frac{1}{6}x^2+o(x^3\ln^4 x).\tag{10} \end{align*}
Now,notice that, by$(8)(10)$, $$f(x)-g(x)=\frac{1}{6}x^2+o(x^3\ln^4 x)\to 0(x \to 0^+),$$ and, by $(10)$, $$\frac{e^g(x)}{x}=e^{g(x)-\ln x}=e^{x\ln^2 x+o(x\ln^2 x)}=e^0=1.$$ It follows that \begin{align*} \lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}&=\lim_{x \to 0+}\frac{e^{f(x)}-e^{g(x)}}{x^3}\\ &=\lim_{x \to 0+}\frac{e^{g(x)}(e^{f(x)-g(x)}-1)}{x^3}\\ &=\lim_{x \to 0+}\frac{e^{g(x)}}{x}\cdot\lim_{x \to 0+}\frac{e^{f(x)-g(x)}-1}{x^2}\\ &=1 \cdot\lim_{x \to 0+}\frac{f(x)-g(x)}{x^2}\\ &=\lim_{x \to 0+}\left(\frac{1}{6}+o(x\ln^4 x)\right)\\ &=\frac{1}{6}. \end{align*}
