Answer
\begin{align}
\operatorname{curl} \mathbf{F}&= -2e^{x} \cos y \ \ \mathbf{k}
\end{align}
\begin{align}
\operatorname{curl} \mathbf{F}(0,0,1)& = -2\mathbf{k}\\
\end{align}
Work Step by Step
Given
$$\mathbf{F}(x, y, z)=e^{x} \sin y \mathbf{i}-e^{x} \cos y \mathbf{j};\ \ \ \ \ (0,0,1) $$
Since \begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}+P\mathbf{k}\end{align}
we get
\begin{array}{l}
{M=e^{x} \sin y } \\
{N= -e^{x} \cos y } \\
{P=0}\end{array}
therefore we have
\begin{align}
\operatorname{curl} \mathbf{F}&=\left|\begin{array}{ccc}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \\ {\frac{\partial}{\partial x}} & {\frac{\partial}{\partial y}} & {\frac{\partial}{\partial z}} \\ {M} & {N} & {P}\end{array}\right|\\
&=\left|\begin{array}{ccc}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \\ {\frac{\partial}{\partial x}} & {\frac{\partial}{\partial y}} & {\frac{\partial}{\partial z}} \\ {e^{x} \sin y } & { -e^{x} \cos y } & {0 }\end{array}\right|\\
&=\left(\frac{\partial}{\partial y}( 0)-\frac{\partial}{\partial z}( -e^{x} \cos y )\right) \mathbf{i}
-\left(\frac{\partial}{\partial x}(0)-\frac{\partial}{\partial z}(e^{x} \sin y )\right) \mathbf{j}
+\left(\frac{\partial}{\partial x}( -e^{x} \cos y )-\frac{\partial}{\partial y}(e^{x} \sin y )\right) \mathbf{k}\\
&=(0-0) \mathbf{i}-(0-0) \mathbf{j}+(-e^{x} \cos y-e^{x} \cos y) \mathbf{k}\\
&=-2e^{x} \cos y \mathbf{k}
\end{align}
\begin{align}
\operatorname{curl} \mathbf{F}(0,0,1)&=-2e^{0} \cos0 \mathbf{k}\\
&=-2 \mathbf{k}
\end{align}
