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