Answer
$\text{The vector field } \mathbf{F} \text{ is not conservative. No potential function } f(x,y,z) \text{ exists.}$
Work Step by Step
\[
\mathbf{F}(x,y,z) = y e^z \mathbf{i} + z e^x \mathbf{j} + x e^y \mathbf{k}
\]
\[
\text{curl } \mathbf{F} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y e^z & z e^x & x e^y
\end{vmatrix}
\]
\[
= \left( \frac{\partial}{\partial y}(x e^y) - \frac{\partial}{\partial z}(z e^x) \right)\mathbf{i}
- \left( \frac{\partial}{\partial x}(x e^y) - \frac{\partial}{\partial z}(y e^z) \right)\mathbf{j}
+ \left( \frac{\partial}{\partial x}(z e^x) - \frac{\partial}{\partial y}(y e^z) \right)\mathbf{k}
\]
\[
= \left( x e^y - e^x \right)\mathbf{i}
- \left( e^y - y e^z \right)\mathbf{j}
+ \left( z e^x - e^z \right)\mathbf{k}
\]
\[
= (x e^y - e^x)\mathbf{i}
+ (y e^z - e^y)\mathbf{j}
+ (z e^x - e^z)\mathbf{k}
\]
\[
\nabla \times \mathbf{F}
= (x e^y - e^x)\mathbf{i} + (y e^z - e^y)\mathbf{j} + (z e^x - e^z)\mathbf{k} \neq \mathbf{0}
\]
