244 已知方程 (dfrac (y)(x),dfrac (z)(x))=0 确定了函数 z=z(x,y),f(u, v)可微,则 dfrac (partial z)(partial x)+ydfrac (partial z)(partial y)=-|||-(A)z. (B) -z. (C)y. (D) -y.A、AB、BC、CD、D

步骤 1：定义变量
设 $u = \dfrac{y}{x}$ 和 $v = \dfrac{z}{x}$，则原方程可以写为 $f(u, v) = 0$。
步骤 2：对x求偏导
对 $f(u, v) = 0$ 关于 $x$ 求偏导，利用链式法则，得到：
$$
\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial u} \cdot \dfrac{\partial u}{\partial x} + \dfrac{\partial f}{\partial v} \cdot \dfrac{\partial v}{\partial x} = 0
$$
其中，$\dfrac{\partial u}{\partial x} = -\dfrac{y}{x^2}$，$\dfrac{\partial v}{\partial x} = \dfrac{1}{x} \cdot \dfrac{\partial z}{\partial x} - \dfrac{z}{x^2}$。
步骤 3：对y求偏导
对 $f(u, v) = 0$ 关于 $y$ 求偏导，利用链式法则，得到：
$$
\dfrac{\partial f}{\partial y} = \dfrac{\partial f}{\partial u} \cdot \dfrac{\partial u}{\partial y} + \dfrac{\partial f}{\partial v} \cdot \dfrac{\partial v}{\partial y} = 0
$$
其中，$\dfrac{\partial u}{\partial y} = \dfrac{1}{x}$，$\dfrac{\partial v}{\partial y} = \dfrac{1}{x} \cdot \dfrac{\partial z}{\partial y}$。
步骤 4：整理方程
将步骤 2 和步骤 3 中的偏导数代入，得到：
$$
\dfrac{\partial f}{\partial u} \cdot \left(-\dfrac{y}{x^2}\right) + \dfrac{\partial f}{\partial v} \cdot \left(\dfrac{1}{x} \cdot \dfrac{\partial z}{\partial x} - \dfrac{z}{x^2}\right) = 0
$$
$$
\dfrac{\partial f}{\partial u} \cdot \dfrac{1}{x} + \dfrac{\partial f}{\partial v} \cdot \dfrac{1}{x} \cdot \dfrac{\partial z}{\partial y} = 0
$$
步骤 5：消去 $\dfrac{\partial f}{\partial u}$ 和 $\dfrac{\partial f}{\partial v}$
将上述两个方程相乘，消去 $\dfrac{\partial f}{\partial u}$ 和 $\dfrac{\partial f}{\partial v}$，得到：
$$
\dfrac{1}{x} \cdot \dfrac{\partial z}{\partial x} - \dfrac{z}{x^2} = -\dfrac{y}{x^2} \cdot \dfrac{\partial z}{\partial y}
$$
步骤 6：整理得到最终结果
将上式整理，得到：
$$
x \cdot \dfrac{\partial z}{\partial x} + y \cdot \dfrac{\partial z}{\partial y} = z
$$