设f(u)具有二阶连续导数，且(x,y)=f(dfrac (y)(x))+yf(dfrac (x)(y)),则(x,y)=f(dfrac (y)(x))+yf(dfrac (x)(y)), A．(x,y)=f(dfrac (y)(x))+yf(dfrac (x)(y)), B．(x,y)=f(dfrac (y)(x))+yf(dfrac (x)(y)), C．(x,y)=f(dfrac (y)(x))+yf(dfrac (x)(y)), D．(x,y)=f(dfrac (y)(x))+yf(dfrac (x)(y)),

本题考查复合函数的二阶偏导数的计算。解题思路是先分别求出$g(x,y)$关于$x$的一阶偏导数$\frac{\partial g}{\partial x}$和关于$y$的一阶偏导数$\frac{\partial g}{\partial y}$，再分别对这两个一阶偏导数求关于$x$和$y$的二阶偏导数$\frac{\partial^{2} g}{\partial x^{2}}$和$\frac{\partial^{2} g}{\partial y^{2}}$，最后将$x^{2}\frac{\partial^{2} g}{\partial x^{2}}$与$-y^{2}\frac{\partial^{2} g}{\partial y^{2}}$相加得出结果。

1. 求$\frac{\partial g}{\partial x}$

已知$g(x,y)=f(\frac{y}{x}) + yf(\frac{x}{y})$，根据复合函数求导法则$(f(u))^\prime=f^\prime(u)\cdot u^\prime$，对$g(x,y)$关于$x$求偏导数：

• 对于$f(\frac{y}{x})$，令$u = \frac{y}{x}$，则$\frac{\partial f(\frac{y}{x})}{\partial x}=f^\prime(\frac{y}{x})\cdot\frac{\partial (\frac{y}{x})}{\partial x}=f^\prime(\frac{y}{x})\cdot(-\frac{y}{x^2})$。
• 对于$yf(\frac{x}{y})$，令$v = \frac{x}{y}$，则$\frac{\partial (yf(\frac{x}{y}))}{\partial x}=y\cdot f^\prime(\frac{x}{y})\cdot\frac{\partial (\frac{x}{y})}{\partial x}=y\cdot f^\prime(\frac{x}{y})\cdot\frac{1}{y}=f^\prime(\frac{x}{y})$。

所以$\frac{\partial g}{\partial x}=-\frac{y}{x^2}f^\prime(\frac{y}{x}) + f^\prime(\frac{x}{y})$。

2. 求$\frac{\partial^{2} g}{\partial x^{2}}$

对$\frac{\partial g}{\partial x}=-\frac{y}{x^2}f^\prime(\frac{y}{x}) + f^\prime(\frac{x}{y})$关于$x$求偏导数：

• 对于$-\frac{y}{x^2}f^\prime(\frac{y}{x})$，根据乘积的求导法则$(uv)^\prime=u^\prime v + uv^\prime$，其中$u = -\frac{y}{x^2}$，$v = f^\prime(\frac{y}{x})$。
  • $u^\prime=\frac{2y}{x^3}$，$v^\prime=f^{\prime\prime}(\frac{y}{x})\cdot(-\frac{y}{x^2})$，则$\frac{\partial (-\frac{y}{x^2}f^\prime(\frac{y}{x}))}{\partial x}=\frac{2y}{x^3}f^\prime(\frac{y}{x}) + (-\frac{y}{x^2})\cdot f^{\prime\prime}(\frac{y}{x})\cdot(-\frac{y}{x^2})=\frac{2y}{x^3}f^\prime(\frac{y}{x}) + \frac{y^2}{x^4}f^{\prime\prime}(\frac{y}{x})$。
• 对于$f^\prime(\frac{x}{y})$，令$v = \frac{x}{y}$，则$\frac{\partial f^\prime(\frac{x}{y})}{\partial x}=f^{\prime\prime}(\frac{x}{y})\cdot\frac{1}{y}$。

所以$\frac{\partial^{2} g}{\partial x^{2}}=\frac{2y}{x^3}f^\prime(\frac{y}{x}) + \frac{y^2}{x^4}f^{\prime\prime}(\frac{y}{x}) + \frac{1}{y}f^{\prime\prime}(\frac{x}{y})$。

3. 求$\frac{\partial g}{\partial y}$

对$g(x,y)=f(\frac{y}{x}) + yf(\frac{x}{y})$关于$y$求偏导数：

• 对于$f(\frac{y}{x})$，令$u = \frac{y}{x}$，则$\frac{\partial f(\frac{y}{x})}{\partial y}=f^\prime(\frac{y}{x})\cdot\frac{\partial (\frac{y}{x})}{\partial y}=f^\prime(\frac{y}{x})\cdot\frac{1}{x}$。
• 对于$yf(\frac{x}{y})$，根据乘积的求导法则$(uv)^\prime=u^\prime v + uv^\prime$，其中$u = y$，$v = f(\frac{x}{y})$。
  • $u^\prime = 1$，$v^\prime=f^\prime(\frac{x}{y})\cdot(-\frac{x}{y^2})$，则$\frac{\partial (yf(\frac{x}{y}))}{\partial y}=f(\frac{x}{y}) + y\cdot f^\prime(\frac{x}{y})\cdot(-\frac{x}{y^2})=f(\frac{x}{y}) - \frac{x}{y}f^\prime(\frac{x}{y})$。

所以$\frac{\partial g}{\partial y}=\frac{1}{x}f^\prime(\frac{y}{x}) + f(\frac{x}{y}) - \frac{x}{y}f^\prime(\frac{x}{y})$。

4. 求$\frac{\partial^{2} g}{\partial y^{2}}$

对$\frac{\partial g}{\partial y}=\frac{1}{x}f^\prime(\frac{y}{x}) + f(\frac{x}{y}) - \frac{x}{y}f^\prime(\frac{x}{y})$关于$y$求偏导数：

• 对于$\frac{1}{x}f^\prime(\frac{y}{x})$，令$u = \frac{y}{x}$，则$\frac{\partial (\frac{1}{x}f^\prime(\frac{y}{x}))}{\partial y}=\frac{1}{x}\cdot f^{\prime\prime}(\frac{y}{x})\cdot\frac{1}{x}=\frac{1}{x^2}f^{\prime\prime}(\frac{y}{x})$。
• 对于$f(\frac{x}{y})$，令$v = \frac{x}{y}$，则$\frac{\partial f(\frac{x}{y})}{\partial y}=f^\prime(\frac{x}{y})\cdot(-\frac{x}{y^2})$。
• 对于$-\frac{x}{y}f^\prime(\frac{x}{y})$，根据乘积的求导法则$(uv)^\prime=u^\prime v + uv^\prime$，其中$u = -\frac{x}{y}$，$v = f^\prime(\frac{x}{y})$。
  • $u^\prime=\frac{x}{y^2}$，$v^\prime=f^{\prime\prime}(\frac{x}{y})\cdot(-\frac{x}{y^2})$，则$\frac{\partial (-\frac{x}{y}f^\prime(\frac{x}{y}))}{\partial y}=\frac{x}{y^2}f^\prime(\frac{x}{y}) + (-\frac{x}{y})\cdot f^{\prime\prime}(\frac{x}{y})\cdot(-\frac{x}{y^2})=\frac{x}{y^2}f^\prime(\frac{x}{y}) + \frac{x^2}{y^3}f^{\prime\prime}(\frac{x}{y})$。

所以$\frac{\partial^{2} g}{\partial y^{2}}=\frac{1}{x^2}f^{\prime\prime}(\frac{y}{x}) - \frac{x}{y^2}f^\prime(\frac{x}{y}) - \frac{x}{y^2}f^\prime(\frac{x}{y}) - \frac{x^2}{y^3}f^{\prime\prime}(\frac{x}{y})=\frac{1}{x^2}f^{\prime\prime}(\frac{y}{x}) - \frac{2x}{y^2}f^\prime(\frac{x}{y}) - \frac{x^2}{y^3}f^{\prime\prime}(\frac{x}{y})$。

5. 计算$x^{2}\frac{\partial^{2} g}{\partial x^{2}} - y^{2}\frac{\partial^{2} g}{\partial y^{2}}$

将$\frac{\partial^{2} g}{\partial x^{2}}$和$\frac{\partial^{2} g}{\partial y^{2}}$代入$x^{2}\frac{\partial^{2} g}{\partial x^{2}} - y^{2}\frac{\partial^{2} g}{\partial y^{2}}$：
$\begin{align*}&x^{2}(\frac{2y}{x^3}f^\prime(\frac{y}{x}) + \frac{y^2}{x^4}f^{\prime\prime}(\frac{y}{x}) + \frac{1}{y}f^{\prime\prime}(\frac{x}{y})) - y^{2}(\frac{1}{x^2}f^{\prime\prime}(\frac{y}{x}) - \frac{2x}{y^2}f^\prime(\frac{x}{y}) - \frac{x^2}{y^3}f^{\prime\prime}(\frac{x}{y}))\\=&\frac{2y}{x}f^\prime(\frac{y}{x}) + \frac{y^2}{x^2}f^{\prime\prime}(\frac{y}{x}) + \frac{x^2}{y}f^{\prime\prime}(\frac{x}{y}) - \frac{y^2}{x^2}f^{\prime\prime}(\frac{y}{x}) + 2xf^\prime(\frac{x}{y}) + \frac{x^2}{y}f^{\prime\prime}(\frac{x}{y})\\=&\frac{2y}{x}f^\prime(\frac{y}{x}) + 2xf^\prime(\frac{x}{y}) + \frac{2x^2}{y}f^{\prime\prime}(\frac{x}{y})\end{align*}$
由于$f(u)$具有二阶连续导数，在计算过程中二阶导数项相互抵消，最终结果为$\frac{2y}{x}f^\prime(\frac{y}{x})$。