5.设 sin (x+2y-3z)=x+2y-3z, 证明: dfrac (partial z)(partial x)+dfrac (partial z)(partial y)=1.

步骤 1：定义函数 $F(x,y,z)$
设 $F(x,y,z)=2\sin (x+2y-3z)-x-2y+3z$，则原方程可以表示为 $F(x,y,z)=0$。

步骤 2：计算偏导数
计算 $F(x,y,z)$ 对 $x$、$y$、$z$ 的偏导数：
- $\dfrac {\partial F}{\partial x}=2\cos (x+2y-3z)-1$
- $\dfrac {\partial F}{\partial y}=4\cos (x+2y-3z)-2$
- $\dfrac {\partial F}{\partial z}=-6\cos (x+2y-3z)+3$

步骤 3：应用隐函数定理
根据隐函数定理，有：
- $\dfrac {\partial z}{\partial x}=-\dfrac {\dfrac {\partial F}{\partial x}}{\dfrac {\partial F}{\partial z}}$
- $\dfrac {\partial z}{\partial y}=-\dfrac {\dfrac {\partial F}{\partial y}}{\dfrac {\partial F}{\partial z}}$

步骤 4：计算 $\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}$
将步骤 2 中的偏导数代入步骤 3 的公式中，得到：
- $\dfrac {\partial z}{\partial x}=-\dfrac {2\cos (x+2y-3z)-1}{-6\cos (x+2y-3z)+3}$
- $\dfrac {\partial z}{\partial y}=-\dfrac {4\cos (x+2y-3z)-2}{-6\cos (x+2y-3z)+3}$

将上述两个表达式相加，得到：
$\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}=-\dfrac {2\cos (x+2y-3z)-1}{-6\cos (x+2y-3z)+3}-\dfrac {4\cos (x+2y-3z)-2}{-6\cos (x+2y-3z)+3}$
$=-\dfrac {2\cos (x+2y-3z)-1+4\cos (x+2y-3z)-2}{-6\cos (x+2y-3z)+3}$
$=-\dfrac {6\cos (x+2y-3z)-3}{-6\cos (x+2y-3z)+3}$
$=1$