【答案】A

【解析】\(z+\ln z - \int_{y}^{x}e^{-t^{2}}dt = 0\)，分别对\(x\)，\(y\)求偏导，得：

\(\frac{\partial z}{\partial x}+\frac{1}{z}\frac{\partial z}{\partial x}-e^{-x^{2}} = 0\Rightarrow\frac{\partial z}{\partial x}=\frac{z}{z + 1}e^{-x^{2}}\)

\(\frac{\partial z}{\partial y}+\frac{1}{z}\frac{\partial z}{\partial y}+e^{-y^{2}} = 0.\Rightarrow\frac{\partial z}{\partial y}=\frac{-z}{z + 1}e^{-y^{2}}\)

\(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=\frac{z}{z + 1}(e^{-x^{2}}-e^{-y^{2}})\)