6.设方程x+z=yf(x²-z²)确定了z=f(x,y)，其中f可微，则z(partial z)/(partial x)+y(partial z)/(partial y)=（） (A.)x. (B.)y. (C.)z. (D.)yf(x²-y²).

为了解决这个问题，我们需要找到由方程 $x + z = yf(x^2 - z^2)$ 确定的函数 $z = z(x, y)$ 的偏导数 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$，然后计算表达式 $z\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y}$。 ### 步骤1：求 $\frac{\partial z}{\partial x}$ 对方程 $x + z = yf(x^2 - z^2)$ 两边关于 $x$ 求偏导数： \[ \frac{\partial}{\partial x}(x + z) = \frac{\partial}{\partial x}(yf(x^2 - z^2)) \] 左边为： \[ 1 + \frac{\partial z}{\partial x} \] 右边使用链式法则，注意到 $f$ 是 $x^2 - z^2$ 的函数： \[ yf'(x^2 - z^2) \cdot \frac{\partial}{\partial x}(x^2 - z^2) = yf'(x^2 - z^2) \cdot (2x - 2z\frac{\partial z}{\partial x}) \] 所以，我们有： \[ 1 + \frac{\partial z}{\partial x} = yf'(x^2 - z^2) \cdot (2x - 2z\frac{\partial z}{\partial x}) \] 整理得： \[ 1 + \frac{\partial z}{\partial x} = 2xyf'(x^2 - z^2) - 2yzf'(x^2 - z^2)\frac{\partial z}{\partial x} \] 将 $\frac{\partial z}{\partial x}$ 项移到一边： \[ \frac{\partial z}{\partial x} + 2yzf'(x^2 - z^2)\frac{\partial z}{\partial x} = 2xyf'(x^2 - z^2) - 1 \] 提取 $\frac{\partial z}{\partial x}$： \[ \frac{\partial z}{\partial x}(1 + 2yzf'(x^2 - z^2)) = 2xyf'(x^2 - z^2) - 1 \] 解得： \[ \frac{\partial z}{\partial x} = \frac{2xyf'(x^2 - z^2) - 1}{1 + 2yzf'(x^2 - z^2)} \] ### 步骤2：求 $\frac{\partial z}{\partial y}$ 对方程 $x + z = yf(x^2 - z^2)$ 两边关于 $y$ 求偏导数： \[ \frac{\partial}{\partial y}(x + z) = \frac{\partial}{\partial y}(yf(x^2 - z^2)) \] 左边为： \[ \frac{\partial z}{\partial y} \] 右边使用乘积法则和链式法则： \[ f(x^2 - z^2) + yf'(x^2 - z^2) \cdot \frac{\partial}{\partial y}(x^2 - z^2) = f(x^2 - z^2) + yf'(x^2 - z^2) \cdot (-2z\frac{\partial z}{\partial y}) \] 所以，我们有： \[ \frac{\partial z}{\partial y} = f(x^2 - z^2) - 2yzf'(x^2 - z^2)\frac{\partial z}{\partial y} \] 将 $\frac{\partial z}{\partial y}$ 项移到一边： \[ \frac{\partial z}{\partial y} + 2yzf'(x^2 - z^2)\frac{\partial z}{\partial y} = f(x^2 - z^2) \] 提取 $\frac{\partial z}{\partial y}$： \[ \frac{\partial z}{\partial y}(1 + 2yzf'(x^2 - z^2)) = f(x^2 - z^2) \] 解得： \[ \frac{\partial z}{\partial y} = \frac{f(x^2 - z^2)}{1 + 2yzf'(x^2 - z^2)} \] ### 步骤3：计算 $z\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y}$ 将 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 代入表达式： \[ z\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = z \cdot \frac{2xyf'(x^2 - z^2) - 1}{1 + 2yzf'(x^2 - z^2)} + y \cdot \frac{f(x^2 - z^2)}{1 + 2yzf'(x^2 - z^2)} \] 合并分数： \[ z\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \frac{z(2xyf'(x^2 - z^2) - 1) + yf(x^2 - z^2)}{1 + 2yzf'(x^2 - z^2)} \] 注意到 $x + z = yf(x^2 - z^2)$，所以 $yf(x^2 - z^2) = x + z$，代入上式： \[ z\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \frac{z(2xyf'(x^2 - z^2) - 1) + x + z}{1 + 2yzf'(x^2 - z^2)} = \frac{2xyzf'(x^2 - z^2) - z + x + z}{1 + 2yzf'(x^2 - z^2)} = \frac{2xyzf'(x^2 - z^2) + x}{1 + 2yzf'(x^2 - z^2)} \] 提取 $x$： \[ z\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \frac{x(1 + 2yzf'(x^2 - z^2))}{1 + 2yzf'(x^2 - z^2)} = x \] 因此，答案是 $\boxed{A}$。