💡 答案解析
方法一 由 $\displaystyle\frac{\partial z}{\partial x}=f_{1}^{\prime}+f_{2}^{\prime}+y f_{3}^{\prime}, \displaystyle\frac{\partial z}{\partial y}=f_{1}^{\prime}-f_{2}^{\prime}+x f_{3}^{\prime}$ ，得
$$
\mathrm{d} z=\frac{\partial z}{\partial x} \mathrm{~d} x+\frac{\partial z}{\partial y} \mathrm{~d} y=\left(f_{1}^{\prime}+f_{2}^{\prime}+y f_{3}^{\prime}\right) \mathrm{d} x+\left(f_{1}^{\prime}-f_{2}^{\prime}+x f_{3}^{\prime}\right) \mathrm{d} y
$$
方法二 $\quad z=f(x+y, x-y, x y)$ 两边求微分得
$$
\begin{gathered}
\mathrm{d} z=f_{1}^{\prime} \cdot \mathrm{d}(x+y)+f_{2}^{\prime} \cdot \mathrm{d}(x-y)+f_{3}^{\prime} \cdot \mathrm{d}(x y) \\
=f_{1}^{\prime} \cdot(\mathrm{d} x+\mathrm{d} y)+f_{2}^{\prime} \cdot(\mathrm{d} x-\mathrm{d} y)+f_{3}^{\prime} \cdot(y \mathrm{~d} x+x \mathrm{~d} y) \\
=\left(f_{1}^{\prime}+f_{2}^{\prime}+y f_{3}^{\prime}\right) \mathrm{d} x+\left(f_{1}^{\prime}-f_{2}^{\prime}+x f_{3}^{\prime}\right) \mathrm{d} y \\
\frac{\partial^{2} z}{\partial x \partial y}=\frac{\partial}{\partial y}\left(f_{1}^{\prime}+f_{2}^{\prime}+y f_{3}^{\prime}\right)=\frac{\partial}{\partial y}\left(f_{1}^{\prime}\right)+\frac{\partial}{\partial y}\left(f_{2}^{\prime}\right)+\frac{\partial}{\partial y}\left(y f_{3}^{\prime}\right) \\
=f_{11}^{\prime \prime}-f_{12}^{\prime \prime}+x f_{13}^{\prime \prime}+f_{21}^{\prime \prime}-f_{22}^{\prime \prime}+x f_{23}^{\prime \prime}+f_{3}^{\prime}+y\left(f_{31}^{\prime \prime}-f_{32}^{\prime \prime}+x f_{33}^{\prime \prime}\right) \\
=f_{11}^{\prime \prime}+(x+y) f_{13}^{\prime \prime}+(x-y) f_{23}^{\prime \prime}-f_{22}^{\prime \prime}+f_{3}^{\prime}+x y f_{33}^{\prime \prime}
\end{gathered}
$$
📋 详细解题步骤
目标：求全微分 dz
设函数 $z = f(u, v, w)$，其中中间变量为 $u = x + y$，$v = x - y$，$w = xy$。根据全微分公式，有 $$dz = \frac{\partial f}{\partial u} du + \frac{\partial f}{\partial v} dv + \frac{\partial f}{\partial w} dw.$$ 分别计算各中间变量的微分：$du = d(x+y) = dx + dy$，$dv = d(x-y) = dx - dy$，$dw = d(xy) = y\,dx + x\,dy$。代入得 $$dz = f_u (dx+dy) + f_v (dx-dy) + f_w (y\,dx + x\,dy).$$ 合并 $dx$ 与 $dy$ 的系数：$dx$ 的系数为 $f_u + f_v + y f_w$，$dy$ 的系数为 $f_u - f_v + x f_w$。因此全微分的表达式为 $$dz = (f_u + f_v + y f_w)\,dx + (f_u - f_v + x f_w)\,dy.$$ 其中 $f_u = \frac{\partial f}{\partial u}$，$f_v = \frac{\partial f}{\partial v}$，$f_w = \frac{\partial f}{\partial w}$ 均为关于 $u, v, w$ 的偏导数，在后续步骤中需进一步代入 $u, v, w$ 的具体表达式。
公式：$$dz = (f_u + f_v + y f_w)\,dx + (f_u - f_v + x f_w)\,dy$$
提示：先写出中间变量的微分，再代入合并系数，注意符号不要遗漏。
目标：求一阶偏导数 ∂z/∂x 和 ∂z/∂y
已知函数 $z = f(u, v, w)$，其中 $u = x + y$，$v = x - y$，$w = xy$。根据复合函数求导法则，$z$ 对 $x$ 和 $y$ 的偏导数需要分别通过中间变量 $u, v, w$ 传递。
首先求 $\frac{\partial z}{\partial x}$。由链式法则：
$$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} + \frac{\partial f}{\partial w} \cdot \frac{\partial w}{\partial x}.$$
计算各中间变量对 $x$ 的偏导数：
$\frac{\partial u}{\partial x} = 1$，$\frac{\partial v}{\partial x} = 1$，$\frac{\partial w}{\partial x} = y$。
代入得：
$$\frac{\partial z}{\partial x} = f_u \cdot 1 + f_v \cdot 1 + f_w \cdot y = f_u + f_v + y f_w.$$
再求 $\frac{\partial z}{\partial y}$。类似地：
$$\frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} + \frac{\partial f}{\partial w} \cdot \frac{\partial w}{\partial y}.$$
计算各中间变量对 $y$ 的偏导数：
$\frac{\partial u}{\partial y} = 1$，$\frac{\partial v}{\partial y} = -1$，$\frac{\partial w}{\partial y} = x$。
代入得：
$$\frac{\partial z}{\partial y} = f_u \cdot 1 + f_v \cdot (-1) + f_w \cdot x = f_u - f_v + x f_w.$$
因此，一阶偏导数的表达式为：
$$\frac{\partial z}{\partial x} = f_u + f_v + y f_w, \quad \frac{\partial z}{\partial y} = f_u - f_v + x f_w.$$
其中 $f_u, f_v, f_w$ 分别表示 $f$ 对第一个、第二个、第三个中间变量的偏导数，它们仍然是 $u, v, w$ 的函数。
公式：\frac{\partial z}{\partial x} = f_u + f_v + y f_w, \quad \frac{\partial z}{\partial y} = f_u - f_v + x f_w
提示：明确中间变量与自变量的关系，逐项应用链式法则，注意符号。
目标：求混合偏导数 ∂²z/∂x∂y
已知 $z = f(u, v, w)$，其中 $u = x^2 + y^2$，$v = xy$，$w = x - y$。第一步已求得 $\frac{\partial z}{\partial x} = f_u \cdot 2x + f_v \cdot y + f_w \cdot 1$。现在对 $\frac{\partial z}{\partial x}$ 关于 $y$ 求偏导，注意 $f_u, f_v, f_w$ 仍是 $u, v, w$ 的函数，而 $u, v, w$ 又依赖于 $x, y$，因此需再次使用链式法则。
对 $\frac{\partial z}{\partial x} = 2x f_u + y f_v + f_w$ 关于 $y$ 求偏导：
$$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}(2x f_u) + \frac{\partial}{\partial y}(y f_v) + \frac{\partial}{\partial y}(f_w).$$
第一项：$\frac{\partial}{\partial y}(2x f_u) = 2x \cdot \frac{\partial f_u}{\partial y}$，而 $f_u$ 是 $u, v, w$ 的函数，故
$$\frac{\partial f_u}{\partial y} = f_{uu} \cdot \frac{\partial u}{\partial y} + f_{uv} \cdot \frac{\partial v}{\partial y} + f_{uw} \cdot \frac{\partial w}{\partial y} = f_{uu} \cdot 2y + f_{uv} \cdot x + f_{uw} \cdot (-1).$$
所以第一项为 $2x(2y f_{uu} + x f_{uv} - f_{uw}) = 4xy f_{uu} + 2x^2 f_{uv} - 2x f_{uw}$。
第二项：$\frac{\partial}{\partial y}(y f_v) = f_v + y \cdot \frac{\partial f_v}{\partial y}$，而
$$\frac{\partial f_v}{\partial y} = f_{vu} \cdot 2y + f_{vv} \cdot x + f_{vw} \cdot (-1).$$
利用 $f$ 的二阶偏导连续性，有 $f_{vu} = f_{uv}$，$f_{vw} = f_{wv}$，故第二项为 $f_v + y(2y f_{uv} + x f_{vv} - f_{wv}) = f_v + 2y^2 f_{uv} + xy f_{vv} - y f_{wv}$。
第三项：$\frac{\partial}{\partial y}(f_w) = \frac{\partial f_w}{\partial y} = f_{wu} \cdot 2y + f_{wv} \cdot x + f_{ww} \cdot (-1) = 2y f_{uw} + x f_{wv} - f_{ww}$（再次利用连续性 $f_{wu}=f_{uw}$）。
将三项相加并合并同类项：
- $f_{uu}$ 项：$4xy f_{uu}$
- $f_{uv}$ 项：$2x^2 f_{uv} + 2y^2 f_{uv} = 2(x^2 + y^2) f_{uv}$
- $f_{uw}$ 项：$-2x f_{uw} + 2y f_{uw} = 2(y - x) f_{uw}$
- $f_{vv}$ 项：$xy f_{vv}$
- $f_{wv}$ 项：$-y f_{wv} + x f_{wv} = (x - y) f_{wv}$
- $f_{ww}$ 项：$- f_{ww}$
- 单独项：$f_v$
因此，最终结果为：
$$\frac{\partial^2 z}{\partial x \partial y} = 4xy f_{uu} + 2(x^2 + y^2) f_{uv} + 2(y - x) f_{uw} + xy f_{vv} + (x - y) f_{wv} - f_{ww} + f_v.$$
注意：$f_{wv}$ 也可写作 $f_{vw}$，$f_{uw}$ 也可写作 $f_{wu}$。此即为所求混合偏导数。
公式：$$\frac{\partial^2 z}{\partial x \partial y} = 4xy f_{uu} + 2(x^2 + y^2) f_{uv} + 2(y - x) f_{uw} + xy f_{vv} + (x - y) f_{wv} - f_{ww} + f_v$$
提示：对每个中间变量分别求导，并利用对称性简化合并同类项。