$$ { 解：}{ 准备工作\left( 求f\prime\left( \xi \right) 的极限 \right) } \\ { 设f\left( x \right) =\left( e^x+x \right) ^{\frac{1}{x}}，首先求其在x\rightarrow 0的极限：} \\ { \lim_{x\rightarrow 0} \left( e^x+x \right) ^{\frac{1}{x}}\begin{array}{l} =\exp \lim_{x\rightarrow 0} \frac{1}{x}\ln \left( e^x+x \right) =\exp \lim_{x\rightarrow 0} \frac{1}{x}\ln \left[ 1+\left( e^x+x-1 \right) \right]\\ =\exp \lim_{x\rightarrow 0} \frac{e^x-1+x}{x}=e^2.\\ \end{array}} \\ { 再对其两边同时取对数有：} \\ { \ln f\left( x \right) \begin{array}{l} =\ln \left( e^x+x \right) ^{\frac{1}{x}}=\frac{\ln \left( e^x+x \right)}{x}=\frac{\ln \left[ e^x\left( 1+\frac{x}{e^x} \right) \right]}{x}=\frac{\ln e^x+\ln \left( 1+\frac{x}{e^x} \right)}{x}\\ =\frac{x+\ln \left( 1+\frac{x}{e^x} \right)}{x}=1+\frac{\ln \left( 1+\frac{x}{e^x} \right)}{x}=1+\frac{\frac{x}{e^x}-\frac{\left( \frac{x}{e^x} \right) ^2}{2}+o\left( x^2 \right)}{x}\\ =1+\frac{\frac{x}{e^x}-\frac{x^2}{2e^{2x}}+o\left( x^2 \right)}{x}=1+\frac{1}{e^x}-\frac{x}{2e^{2x}}+o\left( x \right) .\\ \end{array}} \\ { 设g\left( x \right) =1+\frac{1}{e^x}-\frac{x}{2e^{2x}}，g\left( 0 \right) =2；} \\ { g\prime\left( 0 \right) \begin{array}{l} =\lim_{x\rightarrow 0} \frac{g\left( x \right) -g\left( 0 \right)}{x-0}=\lim_{x\rightarrow 0} \frac{1+\frac{1}{e^x}-\frac{x}{2e^{2x}}-2}{x-0}=\lim_{x\rightarrow 0} \frac{-1+\frac{1}{e^x}-\frac{x}{2e^{2x}}}{x}\\ =\lim_{x\rightarrow 0} \frac{e^{-x}-1}{x}-\lim_{x\rightarrow 0} \frac{1}{2e^{2x}}=\lim_{x\rightarrow 0} \frac{-x}{x}-\frac{1}{2}=-\frac{3}{2}；\\ \end{array}} \\ { 由上述可知，当x\rightarrow 0时，f\left( x \right) \rightarrow e^2，\left[ \ln f\left( x \right) \right] \prime\rightarrow -\frac{3}{2}；} \\ { 对\ln f\left( x \right) 求导：\left[ \ln f\left( x \right) \right] \prime=\frac{1}{f\left( x \right)}\cdot f\prime\left( x \right) \Longrightarrow f\prime\left( x \right) =\ln \left[ \ln f\left( x \right) \right] \prime\cdot f\left( x \right) } \\ { 当x\rightarrow 0时，f\prime\left( x \right) \rightarrow -\frac{3}{2}e^2.} \\ { 求极限：} \\ { \lim_{x\rightarrow 0} \frac{\left( e^{\sin x}+\sin x \right) ^{\frac{1}{\sin x}}-\left( e^{\tan x}+\tan x \right) ^{\frac{1}{\tan x}}}{x^3}} \\ { =\lim_{x\rightarrow 0} \frac{f\left( \sin x \right) -f\left( \tan x \right)}{x^3}{\mathrm{Lagrange}}\lim_{x\rightarrow 0} \frac{f\prime\left( \xi \right) \left( \sin x-\tan x \right)}{x^3}} \\ { =-\frac{3}{2}e^2\cdot \lim_{x\rightarrow 0} \frac{\tan x\left( \cos x-1 \right)}{x^3}=-\frac{3}{2}e^2\cdot \lim_{x\rightarrow 0} \frac{x\cdot \left( -\frac{1}{2}x^2 \right)}{x^3}} \\ { =-\frac{3}{2}e^2\cdot \left( -\frac{1}{2} \right) =\frac{3}{4}e^2.} \\ { 注：拉格朗日中值定理过程} \\ { f\left( \sin x \right) -f\left( \tan x \right) =f\prime\left( \xi \right) \left( \sin x-\tan x \right) ，其中\xi \in \left( \tan x,\sin x \right) } \\ { x\rightarrow 0时，\sin x和\tan x等价于x，即\xi \rightarrow 0；当\xi \rightarrow 0时，f\prime\left( \xi \right) \rightarrow -\frac{3}{2}e^2} $$ 、