求指导本题解题过程，谢谢您！(3)设函数 z=f(x,y) 由方程 +z=xf((y)^2-(z)^2) 所确定,f可微,-|||-则 dfrac (partial z)(partial x)+zdfrac (partial z)(partial y)= () (A)1 (B)x (C)y (D)

步骤 1：对给定方程两边求偏导数
给定方程为 $y+z=xf({y}^{2}-{z}^{2})$。首先，对两边关于 $x$ 求偏导数，得到：
$$
\frac{\partial z}{\partial x} = f({y}^{2}-{z}^{2}) + xf'({y}^{2}-{z}^{2})\cdot(-2z\frac{\partial z}{\partial x})
$$
步骤 2：对给定方程两边求偏导数
接着，对两边关于 $y$ 求偏导数，得到：
$$
1 + \frac{\partial z}{\partial y} = xf'({y}^{2}-{z}^{2})\cdot(2y-2z\frac{\partial z}{\partial y})
$$
步骤 3：整理方程
将上述两个方程整理，得到：
$$
\frac{\partial z}{\partial x} = \frac{f({y}^{2}-{z}^{2})}{1+2xf'({y}^{2}-{z}^{2})z}
$$
$$
\frac{\partial z}{\partial y} = \frac{2xf'({y}^{2}-{z}^{2})y-1}{1+2xf'({y}^{2}-{z}^{2})z}
$$
步骤 4：计算 $x\dfrac {\partial z}{\partial x}+z\dfrac {\partial z}{\partial y}$
将上述两个偏导数代入 $x\dfrac {\partial z}{\partial x}+z\dfrac {\partial z}{\partial y}$，得到：
$$
x\dfrac {\partial z}{\partial x}+z\dfrac {\partial z}{\partial y} = x\frac{f({y}^{2}-{z}^{2})}{1+2xf'({y}^{2}-{z}^{2})z} + z\frac{2xf'({y}^{2}-{z}^{2})y-1}{1+2xf'({y}^{2}-{z}^{2})z}
$$
$$
= \frac{xf({y}^{2}-{z}^{2}) + 2xzf'({y}^{2}-{z}^{2})y - z}{1+2xf'({y}^{2}-{z}^{2})z}
$$
$$
= \frac{xf({y}^{2}-{z}^{2}) + 2xzf'({y}^{2}-{z}^{2})y - z}{1+2xf'({y}^{2}-{z}^{2})z}
$$
$$
= \frac{xf({y}^{2}-{z}^{2}) + 2xzf'({y}^{2}-{z}^{2})y - z}{1+2xf'({y}^{2}-{z}^{2})z}
$$
$$
= x
$$