设=(e)^-(dfrac (1{x)+dfrac (1)(y))}，求证=(e)^-(dfrac (1{x)+dfrac (1)(y))}

考查要点：本题主要考查多元函数的偏导数计算及代数运算能力，需要熟练掌握链式法则的应用。

解题核心思路：

1. 分别计算函数$z$对$x$和$y$的偏导数$\dfrac{\partial z}{\partial x}$和$\dfrac{\partial z}{\partial y}$；
2. 代入表达式$x^2 \dfrac{\partial z}{\partial x} + y^2 \dfrac{\partial z}{\partial y}$，通过化简验证其等于$2z$。

破题关键点：

• 链式法则的正确应用：对复合函数$e^{u(x,y)}$求导时，外层导数为$e^u$，内层导数为$\dfrac{\partial u}{\partial x}$或$\dfrac{\partial u}{\partial y}$；
• 代数化简时注意$x^2$与$\dfrac{1}{x^2}$的抵消关系，最终合并同类项。

步骤1：计算$\dfrac{\partial z}{\partial x}$
设$u = -\left( \dfrac{1}{x} + \dfrac{1}{y} \right)$，则$z = e^u$。
根据链式法则：
$\dfrac{\partial z}{\partial x} = e^u \cdot \dfrac{\partial u}{\partial x} = e^{-\left( \dfrac{1}{x} + \dfrac{1}{y} \right)} \cdot \dfrac{1}{x^2}$

步骤2：计算$\dfrac{\partial z}{\partial y}$
同理，对$y$求偏导：
$\dfrac{\partial z}{\partial y} = e^u \cdot \dfrac{\partial u}{\partial y} = e^{-\left( \dfrac{1}{x} + \dfrac{1}{y} \right)} \cdot \dfrac{1}{y^2}$

步骤3：代入表达式并化简
将偏导数代入$x^2 \dfrac{\partial z}{\partial x} + y^2 \dfrac{\partial z}{\partial y}$：
$\begin{aligned}x^2 \dfrac{\partial z}{\partial x} + y^2 \dfrac{\partial z}{\partial y} &= x^2 \cdot e^{-\left( \dfrac{1}{x} + \dfrac{1}{y} \right)} \cdot \dfrac{1}{x^2} + y^2 \cdot e^{-\left( \dfrac{1}{x} + \dfrac{1}{y} \right)} \cdot \dfrac{1}{y^2} \\&= e^{-\left( \dfrac{1}{x} + \dfrac{1}{y} \right)} + e^{-\left( \dfrac{1}{x} + \dfrac{1}{y} \right)} \\&= 2e^{-\left( \dfrac{1}{x} + \dfrac{1}{y} \right)} = 2z\end{aligned}$