3.设 =dfrac (y)(f({x)^2-(y)^2)} ,其中f为可微函数,验证-|||-dfrac (1)(x)dfrac (partial z)(partial x)+dfrac (1)(y)dfrac (partial z)(partial y)=dfrac (z)({y)^2},

步骤 1：定义中间变量
设 $u = x^2 - y^2$，则 $z = \frac{y}{f(u)}$。
步骤 2：计算偏导数
根据链式法则，计算 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$。
$$
\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} = \frac{-y \cdot f'(u)}{f^2(u)} \cdot 2x = -\frac{2xyf'(u)}{f^2(u)}
$$
$$
\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial y} = \frac{-y \cdot f'(u)}{f^2(u)} \cdot (-2y) + \frac{1}{f(u)} = \frac{2y^2f'(u)}{f^2(u)} + \frac{1}{f(u)}
$$
步骤 3：验证等式
将 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 代入等式 $\frac{1}{x}\frac{\partial z}{\partial x} + \frac{1}{y}\frac{\partial z}{\partial y}$ 中，验证是否等于 $\frac{z}{y^2}$。
$$
\frac{1}{x}\frac{\partial z}{\partial x} + \frac{1}{y}\frac{\partial z}{\partial y} = \frac{1}{x} \cdot \left(-\frac{2xyf'(u)}{f^2(u)}\right) + \frac{1}{y} \cdot \left(\frac{2y^2f'(u)}{f^2(u)} + \frac{1}{f(u)}\right)
$$
$$
= -\frac{2yf'(u)}{f^2(u)} + \frac{2yf'(u)}{f^2(u)} + \frac{1}{yf(u)} = \frac{1}{yf(u)} = \frac{z}{y^2}
$$