11.设 =f((x)^2+(y)^2) ,其中f具有二阶导数,求 dfrac ({a)^2z}(d{x)^2} -dfrac ({partial )^2z}(partial xpartial y) -dfrac ({d)^2z}(partial {y)^2} .

步骤 1：计算 $\dfrac {\partial z}{\partial x}$
根据链式法则，我们有 $\dfrac {\partial z}{\partial x} = \dfrac {df}{d({x}^{2}+{y}^{2})} \cdot \dfrac {\partial ({x}^{2}+{y}^{2})}{\partial x} = 2xf'({x}^{2}+{y}^{2})$。
步骤 2：计算 $\dfrac {\partial z}{\partial y}$
同样地，$\dfrac {\partial z}{\partial y} = \dfrac {df}{d({x}^{2}+{y}^{2})} \cdot \dfrac {\partial ({x}^{2}+{y}^{2})}{\partial y} = 2yf'({x}^{2}+{y}^{2})$。
步骤 3：计算 $\dfrac {{\partial }^{2}z}{\partial {x}^{2}}$
对 $\dfrac {\partial z}{\partial x}$ 再次求导，我们得到 $\dfrac {{\partial }^{2}z}{\partial {x}^{2}} = 2f'({x}^{2}+{y}^{2}) + 4{x}^{2}f''({x}^{2}+{y}^{2})$。
步骤 4：计算 $\dfrac {{\partial }^{2}z}{\partial x\partial y}$
对 $\dfrac {\partial z}{\partial x}$ 关于 $y$ 求导，我们得到 $\dfrac {{\partial }^{2}z}{\partial x\partial y} = 4xyf''({x}^{2}+{y}^{2})$。
步骤 5：计算 $\dfrac {{\partial }^{2}z}{\partial {y}^{2}}$
对 $\dfrac {\partial z}{\partial y}$ 再次求导，我们得到 $\dfrac {{\partial }^{2}z}{\partial {y}^{2}} = 2f'({x}^{2}+{y}^{2}) + 4{y}^{2}f''({x}^{2}+{y}^{2})$。
步骤 6：计算 $\dfrac {{\partial }^{2}z}{\partial {x}^{2}} - \dfrac {{\partial }^{2}z}{\partial x\partial y} - \dfrac {{\partial }^{2}z}{\partial {y}^{2}}$
将上述结果代入，我们得到 $\dfrac {{\partial }^{2}z}{\partial {x}^{2}} - \dfrac {{\partial }^{2}z}{\partial x\partial y} - \dfrac {{\partial }^{2}z}{\partial {y}^{2}} = 2f'({x}^{2}+{y}^{2}) + 4{x}^{2}f''({x}^{2}+{y}^{2}) - 4xyf''({x}^{2}+{y}^{2}) - 2f'({x}^{2}+{y}^{2}) - 4{y}^{2}f''({x}^{2}+{y}^{2})$。
步骤 7：简化表达式
简化上述表达式，我们得到 $\dfrac {{\partial }^{2}z}{\partial {x}^{2}} - \dfrac {{\partial }^{2}z}{\partial x\partial y} - \dfrac {{\partial }^{2}z}{\partial {y}^{2}} = 4({x}^{2} - xy - {y}^{2})f''({x}^{2}+{y}^{2})$。