2.设z=u^2ln v，而u=(x)/(y)，v=3x-2y，求Z_(x)，Z_(y).

设 $ z = u^2 \ln v $，其中 $ u = \frac{x}{y} $，$ v = 3x - 2y $。利用链式法则求偏导数： 1. **计算 $ z_x $** \[ z_x = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} = 2u \ln v \cdot \frac{1}{y} + \frac{u^2}{v} \cdot 3 \] 代入 $ u $ 和 $ v $： \[ z_x = \frac{2x}{y^2} \ln (3x - 2y) + \frac{3x^2}{y^2(3x - 2y)} \] 2. **计算 $ z_y $** \[ z_y = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} = 2u \ln v \cdot \left(-\frac{x}{y^2}\right) + \frac{u^2}{v} \cdot (-2) \] 代入 $ u $ 和 $ v $： \[ z_y = -\frac{2x^2}{y^3} \ln (3x - 2y) - \frac{2x^2}{y^2(3x - 2y)} = -\frac{2x^2}{y^3} \left[ \ln (3x - 2y) + \frac{y}{3x - 2y} \right] \] **答案：** \[ \boxed{ \begin{aligned} z_x &= \frac{2x}{y^2} \ln (3x - 2y) + \frac{3x^2}{y^2(3x - 2y)}, \\ z_y &= -\frac{2x^2}{y^3} \left[ \ln (3x - 2y) + \frac{y}{3x - 2y} \right]. \end{aligned} } \]