18. (10.0分)设函数z=f(x+y,(x)/(y))，其中f具有二阶连续偏导数，求(partial z)/(partial x)，(partial ^2z)/(partial x^2)；

设 $ z = f(u, v) $，其中 $ u = x + y $，$ v = \frac{x}{y} $。

1. 求 $\frac{\partial z}{\partial x}$：
   由链式法则，
   $\frac{\partial z}{\partial x} = f_u \frac{\partial u}{\partial x} + f_v \frac{\partial v}{\partial x} = f_u \cdot 1 + f_v \cdot \frac{1}{y} = f_1' + \frac{1}{y} f_2'$
   其中 $ f_1' = f_u $，$ f_2' = f_v $。

2. 求 $\frac{\partial^2 z}{\partial x^2}$：
   对 $\frac{\partial z}{\partial x}$ 再次关于 $x$ 求偏导，
   $\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left( f_1' + \frac{1}{y} f_2' \right)$
   分别对两项求导：

   • 对 $ f_1' $ 求导：
     $\frac{\partial}{\partial x} (f_1') = f_{11}'' \frac{\partial u}{\partial x} + f_{12}'' \frac{\partial v}{\partial x} = f_{11}'' \cdot 1 + f_{12}'' \cdot \frac{1}{y} = f_{11}'' + \frac{1}{y} f_{12}''$
   • 对 $ \frac{1}{y} f_2' $ 求导：
     $\frac{\partial}{\partial x} \left( \frac{1}{y} f_2' \right) = \frac{1}{y} \left( f_{21}'' \frac{\partial u}{\partial x} + f_{22}'' \frac{\partial v}{\partial x} \right) = \frac{1}{y} \left( f_{21}'' \cdot 1 + f_{22}'' \cdot \frac{1}{y} \right) = \frac{1}{y} f_{21}'' + \frac{1}{y^2} f_{22}''$
     将两部分相加，
     $\frac{\partial^2 z}{\partial x^2} = f_{11}'' + \frac{1}{y} f_{12}'' + \frac{1}{y} f_{21}'' + \frac{1}{y^2} f_{22}'' = f_{11}'' + \frac{2}{y} f_{12}'' + \frac{1}{y^2} f_{22}''$
     （注意 $ f_{12}'' = f_{21}'' $，由连续性）。

最终结果：
$\frac{\partial z}{\partial x} = f_1' + \frac{1}{y} f_2'$
$\frac{\partial^2 z}{\partial x^2} = f_{11}'' + \frac{2}{y} f_{12}'' + \frac{1}{y^2} f_{22}''$