【答案】A。【解析】函数 $z = \sqrt{2 + x^2 + y^2}$，则偏导数为：$$\frac{\partial z}{\partial x} = \frac{x}{\sqrt{2 + x^2 + y^2}},\quad \frac{\partial z}{\partial y} = \frac{y}{\sqrt{2 + x^2 + y^2}}.$$在点 $(1,-1)$ 处，$z = \sqrt{2 + 1 + 1} = \sqrt{4} = 2$，故$$\left.\frac{\partial z}{\partial x}\right|_{(1,-1)} = \frac{1}{2},\quad \left.\frac{\partial z}{\partial y}\right|_{(1,-1)} = \frac{-1}{2}.$$因此梯度 $\operatorname{grad}z(1,-1) = \frac{1}{2}\vec i - \frac{1}{2}\vec j$，对应选项 A。选项 B 符号全反；C、D 偏导数值未除以 $\sqrt{4}=2$，未归一化，错误。