$$ 1 \leftarrow $$ First, compute the gradient of the function $$ p(x, y)=\sqrt{14-3 x^{2}-4 y^{2}} $$. Then evaluate it at the point $$ (-1,-1) $$.

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We start with the function
$$
p(x, y)=\sqrt{14-3 x^{2}-4 y^{2}}.
$$
Let
$$
f(x, y)=14-3x^2-4y^2,
$$
so that
$$
p(x, y)=\sqrt{f(x, y)}.
$$

**Step 1. Compute the partial derivatives using the chain rule.**

For the partial derivative with respect to $$x$$:
$$
\frac{\partial p}{\partial x} = \frac{1}{2\sqrt{f(x,y)}} \cdot \frac{\partial f}{\partial x}.
$$
Since
$$
\frac{\partial f}{\partial x} = -6x,
$$
we get
$$
\frac{\partial p}{\partial x} = \frac{-6x}{2\sqrt{14-3x^2-4y^2}} = \frac{-3x}{\sqrt{14-3x^2-4y^2}}.
$$

Similarly, for the partial derivative with respect to $$y$$:
$$
\frac{\partial p}{\partial y} = \frac{1}{2\sqrt{f(x,y)}} \cdot \frac{\partial f}{\partial y}.
$$
Since
$$
\frac{\partial f}{\partial y} = -8y,
$$
we have
$$
\frac{\partial p}{\partial y} = \frac{-8y}{2\sqrt{14-3x^2-4y^2}} = \frac{-4y}{\sqrt{14-3x^2-4y^2}}.
$$

Thus, the gradient of $$p(x,y)$$ is:
$$
\nabla p(x,y)=\left(\frac{-3x}{\sqrt{14-3x^2-4y^2}},\frac{-4y}{\sqrt{14-3x^2-4y^2}}\right).
$$

**Step 2. Evaluate the gradient at the point $$(-1,-1)$$.**

Substitute $$x=-1$$ and $$y=-1$$ into the expression:
$$
\sqrt{14-3(-1)^2-4(-1)^2} = \sqrt{14-3-4} = \sqrt{7}.
$$

Now, the partial derivatives become:
$$
\frac{\partial p}{\partial x}\Big|_{(-1,-1)} = \frac{-3(-1)}{\sqrt{7}} = \frac{3}{\sqrt{7}},
$$
$$
\frac{\partial p}{\partial y}\Big|_{(-1,-1)} = \frac{-4(-1)}{\sqrt{7}} = \frac{4}{\sqrt{7}}.
$$

**Final Result:**

The gradient of $$p(x,y)$$ at the point $$(-1,-1)$$ is:
$$
\nabla p(-1,-1)=\left(\frac{3}{\sqrt{7}},\frac{4}{\sqrt{7}}\right).
$$
The gradient of $$ p(x, y) $$ at the point $$(-1, -1)$$ is $$ \left( \frac{3}{\sqrt{7}}, \frac{4}{\sqrt{7}} \right) $$.

First, compute the gradient of the function . Then evaluate it at the point .

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