To find a potential function $f(x, y)$ for the vector field $\mathbf{F} = (2y e^{-xy}, 2x e^{-xy})$, we need to identify a scalar function $f(x, y)$ such that

$\mathbf{F} = \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).$

Step-by-Step Solution:

1. Identify the Components: Given $\mathbf{F} = (2y e^{-xy}, 2x e^{-xy})$, let: $F_x = 2y e^{-xy} \quad \text{and} \quad F_y = 2x e^{-xy}.$

2. Integrate $F_x$ with Respect to $x$: To find $f(x, y)$, start by integrating $F_x$ with respect to $x$: $f(x, y) = \int 2y e^{-xy} \, dx.$ Treat $y$ as a constant during this integration: $f(x, y) = -2 e^{-xy} + g(y),$ where $g(y)$ is a function of $y$ alone (since the integration is with respect to $x$).

3. Differentiate with Respect to $y$ and Match with $F_y$: Now, take the partial derivative of $f(x, y)$ with respect to $y$ and set it equal to $F_y$: $\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( -2 e^{-xy} + g(y) \right).$ This gives: $\frac{\partial f}{\partial y} = 2x e^{-xy} + g'(y).$ Set this equal to $F_y = 2x e^{-xy}$: $2x e^{-xy} + g'(y) = 2x e^{-xy}.$ Therefore, $g'(y) = 0$, which implies $g(y)$ is a constant.

4. Conclusion: The potential function $f(x, y)$ is therefore: $f(x, y) = -2 e^{-xy} + C,$ where $C$ is a constant.

Answer:

The correct option is: $\boxed{-2 e^{-xy}}.$

Would you like further details on any part of this solution?

Here are five related questions for further exploration:

1. How can we verify that a given vector field is conservative?
2. What are the necessary conditions for the existence of a potential function?
3. How would the solution change if the vector field had an additional constant component?
4. Can a non-conservative field have a potential function?
5. What is the significance of the constant $C$ in the potential function?

Tip: For conservative vector fields, computing partial derivatives to check equality (mixed partial derivatives) can confirm the existence of a potential function.