Answer
$W$ is not a vector subspace of $R^3$.
Work Step by Step
Let $W=\left\{\left(x_{1}, x_{2}, x_{1} x_{2}\right) : x_{1} \text { and } x_{2} \text { are real numbers }\right\}$, $u=(x_1,x_2,x_1x_2), v=(y_1,y_2,y_1y_2)\in W$. Then
\begin{align*}
u+v&=(x_1,x_2,x_1x_2)+(y_1,y_2,y_1y_2)\\
&= (x_1+y_1,x_2+y_2,x_1x_2+y_1y_2)\\
&\neq (x_1+y_1,x_2+y_2,(x_1+y_1)(x_2+y_2)).\\
\end{align*}
Hence, $u+v\not\in W$ and $W$ is not vector subspace of $R^3$.