The problem asks us to find the equation of the tangent plane to the surface $z = xe^{-2y}$ at the point $(1, 0, 1)$.

1. Problem Description

The problem asks us to find the equation of the tangent plane to the surface $z = xe^{-2y}$ at the point $(1, 0, 1)$.

2. Solution Steps

First, we need to find the partial derivatives of $z$ with respect to $x$ and $y$.

$\frac{\partial z}{\partial x} = e^{-2y}$

$\frac{\partial z}{\partial y} = x(-2e^{-2y}) = -2xe^{-2y}$

Next, we evaluate these partial derivatives at the given point $(1, 0, 1)$.

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

$\frac{\partial z}{\partial y}(1, 0) = -2(1)e^{-2(0)} = -2(1)(1) = -2$

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

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

In this case, $(x_0, y_0, z_0) = (1, 0, 1)$. So the equation of the tangent plane is:

$z - 1 = 1(x - 1) + (-2)(y - 0)$

$z - 1 = x - 1 - 2y$

$z = x - 2y$

3. Final Answer

$z = x - 2y$

or

$x - 2y - z = 0$