$$ { 解：令u=xy+z-t，则t=xy+z-u，dt=\mathrm{d}\left( xy+z-u \right) =-\mathrm{d}u；} \\ { 积分上下限为：u_{上}=xy+z-t_{上}=xy+z-z=xy；u_{下}=xy+z-t_{下}=xy+z-xy=z；} \\ { \int_{xy}^z{g\left( xy+z-t \right)}\mathrm{d}t=\int_z^{xy}{g\left( u \right)}\left( -\mathrm{d}u \right) =\int_{xy}^z{g\left( u \right)}\mathrm{d}u=\int_{xy}^z{g\left( t \right)}\mathrm{d}t；} \\ { 令G\left( x,y,z \right) =\int_{xy}^z{g\left( t \right)}\mathrm{d}t-\mathrm{e}^{xyz}，则有：\begin{cases} G_x=-yg\left( xy \right) -yz\mathrm{e}^{xyz}\\ G_y=-xg\left( xy \right) -xz\mathrm{e}^{xyz}\\ G_z=g\left( xy \right) -z\mathrm{e}^{xyz}\\ \end{cases}；} \\ { 根据隐函数存在定理求解\frac{\partial z}{\partial x}和\frac{\partial z}{\partial y}：} \\ { \begin{cases} \frac{\partial z}{\partial x}=-\frac{G_x}{G_z}=-\frac{-yg\left( xy \right) -yz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}}=\frac{yg\left( xy \right) +yz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}}\\ \frac{\partial z}{\partial y}=-\frac{G_y}{G_z}=-\frac{-xg\left( xy \right) -xz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}}=\frac{xg\left( xy \right) +xz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}}\\ \end{cases}；} \\ { 下面求解\frac{\partial u}{\partial x}和\frac{\partial u}{\partial y}：} \\ { \begin{cases} \frac{\partial u}{\partial x}=f_{1}^{\prime}+f_{3}^{\prime}\cdot \left( yz+\frac{\partial z}{\partial x} \right) =f_{1}^{\prime}+f_{3}^{\prime}\cdot \left[ yz+\frac{yg\left( xy \right) +yz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right]\\ \frac{\partial u}{\partial y}=f_{2}^{\prime}+f_{3}^{\prime}\cdot \left( xz+\frac{\partial z}{\partial y} \right) =f_{2}^{\prime}+f_{3}^{\prime}\cdot \left[ xz+\frac{xg\left( xy \right) +xz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right]\\ \end{cases}；} \\ { 最后求解x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}：} \\ { x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}\begin{array}{l} =x\cdot \left\{ f_{1}^{\prime}+f_{3}^{\prime}\cdot \left[ yz+\frac{yg\left( xy \right) +yz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right] \right\} -y\cdot \left[ f_{2}^{\prime}+f_{3}^{\prime}\cdot \left[ xz+\frac{xg\left( xy \right) +xz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right] \right]\\ =xf_{1}^{\prime}-yf_{2}^{\prime}+f_{3}^{\prime}\left[ xyz+\frac{xyg\left( xy \right) +xyz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right] -f_{3}^{\prime}\left[ xyz+\frac{xyg\left( xy \right) +xyz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right]\\ =xf_{1}^{\prime}-yf_{2}^{\prime}.\\ \end{array}} \\ $$