(3)已知函数z=f(xy,e^x+y)，且f(x,y)具有二阶连续偏导数.则(partial^2z)/(partial xpartial y)=_____.

设 $u = xy$，$v = e^{x+y}$，则 $z = f(u, v)$。 一阶偏导数： \[ \frac{\partial z}{\partial x} = f_1 \cdot y + f_2 \cdot e^{x+y} \] 二阶混合偏导数： \[ \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y} \left( f_1 \cdot y + f_2 \cdot e^{x+y} \right) = f_1 + y \frac{\partial f_1}{\partial y} + f_2 e^{x+y} + e^{x+y} \frac{\partial f_2}{\partial y} \] 利用链式法则： \[ \frac{\partial f_1}{\partial y} = f_{11} \cdot x + f_{12} \cdot e^{x+y}, \quad \frac{\partial f_2}{\partial y} = f_{21} \cdot x + f_{22} \cdot e^{x+y} \] 代入得： \[ \frac{\partial^2 z}{\partial x \partial y} = f_1 + xy f_{11} + (x + y) e^{x+y} f_{12} + e^{x+y} f_2 + e^{2x+2y} f_{22} \] **答案：** \[ \boxed{f_1 + xy f_{11} + (x + y) e^{x+y} f_{12} + e^{x+y} f_2 + e^{2x+2y} f_{22}} \]