Answer
The vector field is not conservative
Work Step by Step
$F(x, y, z) = \sin y\mathbf{i} - x\cos y\mathbf{j} + \mathbf{k}$
$\text{curl } \mathbf{F} = \nabla \times \mathbf{F} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
P & Q & R
\end{vmatrix}$
where $P = \sin y$, $Q = -x\cos y$, and $R = 1$.
$\text{curl } \mathbf{F} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\sin y & -x\cos y & 1
\end{vmatrix}$
$\text{curl } \mathbf{F} = \left( \frac{\partial}{\partial y}(1) - \frac{\partial}{\partial z}(-x\cos y) \right)\mathbf{i} - \left( \frac{\partial}{\partial x}(1) - \frac{\partial}{\partial z}(\sin y) \right)\mathbf{j} + \left( \frac{\partial}{\partial x}(-x\cos y) - \frac{\partial}{\partial y}(\sin y) \right)\mathbf{k}$
$= \left( 0 - 0 \right)\mathbf{i} - \left( 0 - 0 \right)\mathbf{j} + \left( -\cos y - \cos y \right)\mathbf{k}$
$= 0\mathbf{i} - 0\mathbf{j} + (-2\cos y)\mathbf{k}$
$= -2\cos y\mathbf{k}$
Since $\text{curl } \mathbf{F} = -2\cos y\mathbf{k} \neq \mathbf{0}$, the vector field is not conservative.
