Answer
\begin{align}
\operatorname{curl} \mathbf{F}&=(z+2x) \mathbf{i}+x^2 \mathbf{j} -2z \mathbf{k}
\end{align}
\begin{align}
\operatorname{curl} \mathbf{F}(2,1,3)&=7 \mathbf{i}+4 \mathbf{j}-6\mathbf{k}\\
\end{align}
Work Step by Step
Given
$$\mathbf{F}(x, y, z)=\mathbf{F}(x, y, z)=x^{2} z \mathbf{i}-2 x z \mathbf{j}+y z \mathbf{k} ;\ \ \ \ \ (2,-1,3) $$
Since \begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}+P\mathbf{k}\end{align}
we get
\begin{array}{l}
{M=x^2 z} \\
{N=-2xz} \\
{P=yz}\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^2 z} & {-2x z} & { y z}\end{array}\right|\\
&=\left(\frac{\partial}{\partial y}( yz)-\frac{\partial}{\partial z}(-2x z)\right) \mathbf{i}
-\left(\frac{\partial}{\partial x}(y z)-\frac{\partial}{\partial z}(x^2 z)\right) \mathbf{j}
+\left(\frac{\partial}{\partial x}(-2x z)-\frac{\partial}{\partial y}(x^2 z)\right) \mathbf{k}\\
&=(z+2x) \mathbf{i}-(0-x^2) \mathbf{j}+(-2z-0) \mathbf{k}\\
&=(z+2x) \mathbf{i}+x^2 \mathbf{j} -2z \mathbf{k}
\end{align}
\begin{align}
\operatorname{curl} \mathbf{F}(2,-1,3)&=(3+4) \mathbf{i}+2^2 \mathbf{j} -6 \mathbf{k}\\
&=7 \mathbf{i}+4 \mathbf{j}-6\mathbf{k}\\
\end{align}
