\( 1 \leftarrow \quad \) First, compute the gradient of the function \( p(x, y)=\sqrt{10-3 x^{2}-y^{2}} \). Then evaluate it at the point \( (-1,-1) \). The gradient is \( \nabla p(x, y)=\langle\square, \square \).

We start with the function \[ p(x,y)=\sqrt{10-3x^2-y^2}. \] Define \[ u=10-3x^2-y^2, \] so that \[ p(x,y)=\sqrt{u}. \] **Step 1. Compute the partial derivative with respect to \( x \):** Using the chain rule, \[ \frac{\partial p}{\partial x}=\frac{1}{2\sqrt{u}}\,\frac{\partial u}{\partial x}. \] Since \[ \frac{\partial u}{\partial x}=-6x, \] we have \[ \frac{\partial p}{\partial x}=\frac{-6x}{2\sqrt{10-3x^2-y^2}}=-\frac{3x}{\sqrt{10-3x^2-y^2}}. \] **Step 2. Compute the partial derivative with respect to \( y \):** Similarly, \[ \frac{\partial p}{\partial y}=\frac{1}{2\sqrt{u}}\,\frac{\partial u}{\partial y}. \] Since \[ \frac{\partial u}{\partial y}=-2y, \] we obtain \[ \frac{\partial p}{\partial y}=\frac{-2y}{2\sqrt{10-3x^2-y^2}}=-\frac{y}{\sqrt{10-3x^2-y^2}}. \] Thus, the gradient of \( p(x,y) \) is \[ \nabla p(x,y)=\left\langle -\frac{3x}{\sqrt{10-3x^2-y^2}},\; -\frac{y}{\sqrt{10-3x^2-y^2}} \right\rangle. \] **Step 3. Evaluate the gradient at the point \( (-1,-1) \):** First, compute the denominator: \[ \sqrt{10-3(-1)^2-(-1)^2}=\sqrt{10-3-1}=\sqrt{6}. \] Now, plug in \( x=-1 \) and \( y=-1 \): For the \( x \)-component: \[ -\frac{3(-1)}{\sqrt{6}}=\frac{3}{\sqrt{6}}. \] For the \( y \)-component: \[ -\frac{-1}{\sqrt{6}}=\frac{1}{\sqrt{6}}. \] Therefore, the gradient evaluated at \( (-1,-1) \) is \[ \nabla p(-1,-1)=\left\langle \frac{3}{\sqrt{6}},\; \frac{1}{\sqrt{6}} \right\rangle. \] Thus, the completed answer is: \[ \nabla p(x,y)=\left\langle -\frac{3x}{\sqrt{10-3x^2-y^2}},\; -\frac{y}{\sqrt{10-3x^2-y^2}} \right\rangle \] and \[ \nabla p(-1,-1)=\left\langle \frac{3}{\sqrt{6}},\; \frac{1}{\sqrt{6}} \right\rangle. \]