###9. 设$z=xy+xF(u),u=\frac y x$,$F(u)$可导，证明
$x\frac{\partial z}{\partial x}+y\frac{\partial z}{y}=z+xy$

###12(3) 求函$z=f(xy^2,x^2y)$数的$\frac{\partial^2z}{\partial x^2}$,$\frac{\partial^2 z}{\partial x\partial y}$, $\frac{\partial^2z}{\partial y^2}$.
#### 解
$\frac{\partial z}{\partial x}=f^{'}_1y^2+f^{'}_22xy$

$\frac{\partial z}{\partial y}=f^{'}_12xy+f^{'}_2x^2$

- $\frac {\partial^2 z}{\partial x^2}=\frac{\partial {\frac{\partial z}{\partial x}}}{\partial x}=y^2 \frac{\partial f^{'}_1}{\partial x}+2yf^{'}_2+\frac{\partial f^{'}_2}{\partial x}2xy$

$=y^2 (f^{''}_{11}y^2+f^{''}_{12}2xy)+2yf^{'}_2+(f^{''}_{21}y^2+f^{''}_{22}2xy)2xy$

$=2yf{'}_2+y^4f^"_{11}+4xy^3f^"_{12}+4x^2y^2f^"_{22}$

- $\frac {\partial^2 z}{\partial y^2}=\frac{\partial {\frac{\partial z}{\partial y}}}{\partial y}=\frac{\partial f^{'}_1}{\partial y}2xy+2xf^{'}_1+\frac{\partial f^{'}_2}{\partial y}x^2$

$=(f^{''}_{11}2xy+f^{''}_{12}x^2)2xy+2xf^{'}_1+x^2 (f^{''}_{21}2xy+f^{''}_{22}x^2)$

$=2xf^{'}_1+4x^2y^2f^"_{11}+4x^3yf^"_{12}+x^4f^"_{22}$

- $\frac{\partial^2z}{\partial x\partial y}=\frac{\partial(\frac{\partial z}{\partial x})}{\partial y}=\frac{\partial f^{'}_1}{\partial y}y^2+2y f^{'}_1+\frac{\partial f^{'}_2}{\partial y}2xy+2x f^{'}_2$

$=2y f^{'}_1+2x f^{'}_2+(f^"_{11}2xy+f^"_{12}x^2)y^2+(f^"_{21}2xy+f^"_{22}x^2)2xy$

$=2y f^{'}_1+2xf^{'}_2+2xy^3f^"_{11}+5x^2y^2f^"_{12}+2x^3yf^"_{22}$