Answer
\begin{align}
\operatorname{curl} \mathbf{F}&=(x z-x y) \mathbf{i}-(y z-x y) \mathbf{j}+(y z-x z) \mathbf{k}
\end{align}
\begin{align}
\operatorname{curl} \mathbf{F}(2,1,3)& =4 \mathbf{i}- \mathbf{j}-3\mathbf{k}\\
\end{align}
Work Step by Step
Given
$$\mathbf{F}(x, y, z)=x y z \mathbf{i}+x y z \mathbf{j}+x y z \mathbf{k} ;\ \ \ \ \ (2,1,3) $$
Since \begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}+P\mathbf{k}\end{align}
we get
\begin{array}{l}
{M=xyz} \\
{N=xyz} \\
{P=xyz}\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}} \\ {x y z} & {x y z} & {x y z}\end{array}\right|\\
&=\left(\frac{\partial}{\partial y}(x yz)-\frac{\partial}{\partial z}(x yz)\right) \mathbf{i}
-\left(\frac{\partial}{\partial x}(xy z)-\frac{\partial}{\partial z}(x yz)\right) \mathbf{j}
+\left(\frac{\partial}{\partial x}(xy z)-\frac{\partial}{\partial y}(x yz)\right) \mathbf{k}\\
&=(x z-x y) \mathbf{i}-(y z-x y) \mathbf{j}+(y z-x z) \mathbf{k}
\end{align}
\begin{align}
\operatorname{curl} \mathbf{F}(2,1,3)&=(6-2) \mathbf{i}-(3-2) \mathbf{j}+(3-6) \mathbf{k}\\
&=4 \mathbf{i}- \mathbf{j}-3\mathbf{k}\\
\end{align}
