4.设 z=xy+f(u) ,, =dfrac (y)(x) ,f(u)为可微函数,求: dfrac (partial z)(partial x)+ydfrac (partial z)(partial y)

步骤 1：计算 $\dfrac {\partial z}{\partial x}$
根据题目，$z=xy+f(u)$，其中 $u=\dfrac {y}{x}$。首先，我们需要计算 $z$ 对 $x$ 的偏导数。根据链式法则，我们有：
$$\dfrac {\partial z}{\partial x} = \dfrac {\partial (xy)}{\partial x} + \dfrac {\partial f(u)}{\partial x}$$
$$= y + f'(u) \cdot \dfrac {\partial u}{\partial x}$$
$$= y + f'(u) \cdot \left(-\dfrac {y}{{x}^{2}}\right)$$
$$= y - \dfrac {y}{x}f'(u)$$
步骤 2：计算 $\dfrac {\partial z}{\partial y}$
接下来，我们需要计算 $z$ 对 $y$ 的偏导数。同样地，根据链式法则，我们有：
$$\dfrac {\partial z}{\partial y} = \dfrac {\partial (xy)}{\partial y} + \dfrac {\partial f(u)}{\partial y}$$
$$= x + f'(u) \cdot \dfrac {\partial u}{\partial y}$$
$$= x + f'(u) \cdot \dfrac {1}{x}$$
$$= x + \dfrac {1}{x}f'(u)$$
步骤 3：计算 $x\dfrac {\partial z}{\partial x}+y\dfrac {\partial z}{\partial y}$
最后，我们需要计算 $x\dfrac {\partial z}{\partial x}+y\dfrac {\partial z}{\partial y}$。将上面计算的偏导数代入，我们有：
$$x\dfrac {\partial z}{\partial x}+y\dfrac {\partial z}{\partial y} = x\left(y - \dfrac {y}{x}f'(u)\right) + y\left(x + \dfrac {1}{x}f'(u)\right)$$
$$= xy - yf'(u) + xy + yf'(u)$$
$$= 2xy$$