Let's solve each part step by step to determine if the vector fields are conservative and, if they are, find the potential function $f(x, y)$.

Part A:

$\mathbf{F}(x, y) = (-4x + 2y)\mathbf{i} + (2x + 4y)\mathbf{j}$

To check if $\mathbf{F}$ is conservative:

1. Compute the partial derivatives:

   • $\frac{\partial P}{\partial y}$, where $P = -4x + 2y$.
   • $\frac{\partial Q}{\partial x}$, where $Q = 2x + 4y$.

   $\frac{\partial P}{\partial y} = 2, \quad \frac{\partial Q}{\partial x} = 2$

   Since $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, the field is conservative.

2. Find the potential function $f(x, y)$:

   • Integrate $P = -4x + 2y$ with respect to $x$: $f(x, y) = \int (-4x + 2y) \, dx = -2x^2 + 2xy + C(y),$ where $C(y)$ is an arbitrary function of $y$.

   • Differentiate $f(x, y)$ with respect to $y$ and match it to $Q$: $\frac{\partial f}{\partial y} = 2x + C'(y) = 2x + 4y$

     From this, $C'(y) = 4y$, so $C(y) = 2y^2$.

     Thus, the potential function is: $f(x, y) = -2x^2 + 2xy + 2y^2$

Part B:

$\mathbf{F}(x, y) = -2y\mathbf{i} - x\mathbf{j}$

To check if $\mathbf{F}$ is conservative:

1. Compute the partial derivatives:

   • $\frac{\partial P}{\partial y}$, where $P = -2y$.
   • $\frac{\partial Q}{\partial x}$, where $Q = -x$.

   $\frac{\partial P}{\partial y} = -2, \quad \frac{\partial Q}{\partial x} = -1$

   Since $\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}$, the field is not conservative.

   Answer: $f(x, y) = N$.

Part C:

$\mathbf{F}(x, y) = (-2\sin y)\mathbf{i} + (4y - 2x\cos y)\mathbf{j}$

To check if $\mathbf{F}$ is conservative:

1. Compute the partial derivatives:

   • $\frac{\partial P}{\partial y}$, where $P = -2\sin y$.
   • $\frac{\partial Q}{\partial x}$, where $Q = 4y - 2x\cos y$.

   $\frac{\partial P}{\partial y} = -2\cos y, \quad \frac{\partial Q}{\partial x} = -2\cos y$

   Since $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, the field is conservative.

2. Find the potential function $f(x, y)$:

   • Integrate $P = -2\sin y$ with respect to $x$: $f(x, y) = \int (-2\sin y) \, dx = -2x\sin y + C(y),$ where $C(y)$ is an arbitrary function of $y$.

   • Differentiate $f(x, y)$ with respect to $y$ and match it to $Q$: $\frac{\partial f}{\partial y} = -2x\cos y + C'(y) = 4y - 2x\cos y$

     From this, $C'(y) = 4y$, so $C(y) = 2y^2$.

     Thus, the potential function is: $f(x, y) = -2x\sin y + 2y^2$

Final Answers:

1. A: $f(x, y) = -2x^2 + 2xy + 2y^2$
2. B: $f(x, y) = N$
3. C: $f(x, y) = -2x\sin y + 2y^2$

Do you have any questions or need further clarification? Here are some related questions to expand your understanding:

1. How do you check if a vector field is conservative in three dimensions?
2. What is the significance of the potential function in physics or engineering?
3. Can every vector field have a potential function? Why or why not?
4. What happens if the vector field is defined over a non-simply connected domain?
5. How do line integrals simplify for conservative vector fields?

Tip: Always double-check the mixed partial derivatives' equality ($\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$) when determining conservativeness.