Answer
\begin{align}
\operatorname{curl} \mathbf{F}&=-x(z-y) e^{-x y z}\mathbf{i}
+y(z-x) e^{-x y z}\mathbf{j}
-z(y -x) e^{-x y z}\mathbf{k}
\end{align}
\begin{align}
\operatorname{curl} \mathbf{F}(3,2,0)& =6 \mathbf{i}-6 \mathbf{j} \\
\end{align}
Work Step by Step
Given
$$\mathbf{F}(x, y, z)=e^{-x y z}(\mathbf{i}+ \mathbf{j}+ \mathbf{k})\\
=e^{-x y z} \mathbf{i}+e^{-x y z} \mathbf{j}+e^{-x y z} \mathbf{k} ;\ \ \ \ \ (3,2,0) $$
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 y z}} \\
{N=e^{-x y z}} \\
{P=e^{-x y z}}\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 y z}} & {e^{-x y z}} & {e^{-x y z}}\end{array}\right|\\
&=\left(\frac{\partial}{\partial y}(e^{-x y z})-\frac{\partial}{\partial z}(xe^{-x y z})\right) \mathbf{i}\\
&
\quad-\left(\frac{\partial}{\partial x}(e^{-x y z})-\frac{\partial}{\partial z}(e^{-x y z})\right) \mathbf{j}\\
&
\quad+\left(\frac{\partial}{\partial x}(e^{-x y z})-\frac{\partial}{\partial y}(e^{-x y z})\right) \mathbf{k}\\
&=(-xze^{-x y z}-(-xy)e^{-x y z}) \mathbf{i}\\
&\quad-(-yze^{-x y z}-(-xy)e^{-x y z}) \mathbf{j}\\
&\quad+(-yze^{-x y z}-(-xz)e^{-x y z}) \mathbf{k}\\
&=(-xz+xy) e^{-x y z}\mathbf{i}
-(-yz+xy) e^{-x y z}\mathbf{j}
+(-yz +xz) e^{-x y z}\mathbf{k}\\
&=-x(z-y) e^{-x y z}\mathbf{i}
+y(z-x) e^{-x y z}\mathbf{j}
-z(y -x) e^{-x y z}\mathbf{k}
\end{align}
\begin{align}
\operatorname{curl} \mathbf{F}(3,2,0)
&=-x(z-y) e^{-x y z}\mathbf{i}
+y(z-x) e^{-x y z}\mathbf{j}
-z(y -x) e^{-x y z}\mathbf{k}
&=4 \mathbf{i}- \mathbf{j}-3\mathbf{k}\\
&=-3(0-2) e^{0}\mathbf{i}
+2(0-3) e^{0}\mathbf{j}
-0(2 -3) e^{0}\mathbf{k} \\
&=6 \mathbf{i}
-6 \mathbf{j}
\end{align}
