1. Let $x$ and $y$ be real numbers that satisfy the following system of equations:
$$
\left\{\begin{array}{l}
\frac{x}{x^{2} y^{2}-1}-\frac{1}{x}=4 \\
\frac{x^{2} y}{x^{2} y^{2}-1}+y=2
\end{array}\right.
$$

Find all possible values of the product $x y$.
1. The given system can be expressed as follows:
$$
\left\{\begin{array} { l } 
{ \frac { x } { x ^ { 2 } y ^ { 2 } - 1 } - \frac { 1 } { x } = 4 } \\
{ \frac { x ^ { 2 } y } { x ^ { 2 } y ^ { 2 } - 1 } + y = 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
\frac{x^{2}}{x^{2} y^{2}-1}-1=4 x \\
\frac{x^{2}}{x^{2} y^{2}-1}+1=\frac{2}{y}
\end{array}\right.\right.
$$

We then have
$$
4 x+\frac{2}{y}=\frac{2 x^{2}}{x^{2} y^{2}-1} \Rightarrow 2 x+\frac{1}{y}=\frac{x^{2}}{x^{2} y^{2}-1}
$$
and
$$
4 x-\frac{2}{y}=-2 \Rightarrow 2 x-\frac{1}{y}=-1 \text {, }
$$
which gives us
$$
\begin{aligned}
\left(2 x+\frac{1}{y}\right)\left(2 x-\frac{1}{y}\right) & =\frac{-x^{2}}{x^{2} y^{2}-1} \\
4 x^{2}-\frac{1}{y^{2}} & =\frac{-x^{2}}{x^{2} y^{2}-1} \\
\frac{4 x^{2} y^{2}-1}{y^{2}} & =\frac{-x^{2}}{x^{2} y^{2}-1} \\
\left(4 x^{2} y^{2}-1\right)\left(x^{2} y^{2}-1\right) & =-x^{2} y^{2} \\
4 x^{4} y^{4}-5 x^{2} y^{2}+1 & =-x^{2} y^{2} \\
4 x^{4} y^{4}-4 x^{2} y^{2}+1 & =0 \\
\left(2 x^{2} y^{2}-1\right)^{2} & =0 \Rightarrow x^{2} y^{2}=\frac{1}{2} \Rightarrow x y= \pm \frac{1}{\sqrt{2}}
\end{aligned}
$$