We are given the function $f(x, y) = \frac{x^2}{y}$ and the point $p = (2, -1)$. We need to find the gradient vector of $f$ at $p$ and the equation of the tangent plane to the surface $z = f(x, y)$ at $p$.

1. Problem Description

We are given the function $f(x, y) = \frac{x^2}{y}$ and the point $p = (2, -1)$. We need to find the gradient vector of $f$ at $p$ and the equation of the tangent plane to the surface $z = f(x, y)$ at $p$.

2. Solution Steps

First, we compute the partial derivatives of $f(x, y)$ with respect to $x$ and $y$:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (\frac{x^2}{y}) = \frac{2x}{y}$

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (\frac{x^2}{y}) = x^2 \frac{\partial}{\partial y} (\frac{1}{y}) = x^2 (-\frac{1}{y^2}) = -\frac{x^2}{y^2}$

Next, we evaluate the partial derivatives at the point $p = (2, -1)$:

$\frac{\partial f}{\partial x}(2, -1) = \frac{2(2)}{-1} = -4$

$\frac{\partial f}{\partial y}(2, -1) = -\frac{(2)^2}{(-1)^2} = -\frac{4}{1} = -4$

The gradient vector at the point $p$ is:

$\nabla f(2, -1) = (\frac{\partial f}{\partial x}(2, -1), \frac{\partial f}{\partial y}(2, -1)) = (-4, -4)$

Now, we find the $z$ value at the point $p = (2, -1)$:

$z_0 = f(2, -1) = \frac{(2)^2}{-1} = \frac{4}{-1} = -4$

Thus, the point on the surface is $(2, -1, -4)$.

The equation of the tangent plane at the point $(x_0, y_0, z_0)$ is given by:

$z - z_0 = \frac{\partial f}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0, y_0)(y - y_0)$

Plugging in the values we found, we get:

$z - (-4) = -4(x - 2) - 4(y - (-1))$

$z + 4 = -4x + 8 - 4y - 4$

$z = -4x - 4y + 8 - 4 - 4$

$z = -4x - 4y$

$4x + 4y + z = 0$

3. Final Answer

The gradient vector is $\nabla f(2, -1) = (-4, -4)$.

The equation of the tangent plane at $p$ is $4x + 4y + z = 0$.