z=lne^x+yx=uvy=u-v求zu与zv

首先，根据链式法则，有：

$$ \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial u} $$

$$ \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial v} $$

接下来，我们分别计算 $\frac{\partial z}{\partial x}$、$\frac{\partial z}{\partial y}$、$\frac{\partial x}{\partial u}$、$\frac{\partial x}{\partial v}$、$\frac{\partial y}{\partial u}$ 和 $\frac{\partial y}{\partial v}$。

$$ \frac{\partial z}{\partial x} = \frac{1}{e^x+y} \cdot e^x = \frac{e^x}{e^x+y} $$

$$ \frac{\partial z}{\partial y} = \frac{1}{e^x+y} $$

$$ \frac{\partial x}{\partial u} = v $$

$$ \frac{\partial x}{\partial v} = u $$

$$ \frac{\partial y}{\partial u} = 1 $$

$$ \frac{\partial y}{\partial v} = -1 $$

将上述结果代入链式法则中，得到：

$$ \frac{\partial z}{\partial u} = \frac{e^u}{e^u-2v+y} $$

$$ \frac{\partial z}{\partial v} = \frac{-e^u}{e^u-2v+y} $$

因此，$zu=\frac{e^u}{e^u-2v+y}$，$zv=\frac{-e^u}{e^u-2v+y}$