Find all the $x$-coordinates (in increasing order) of the points on the curve $x^{2} y^{2}+x y=2$ where the slope of the tangent line is -1 .
\[
\begin{array}{l}
x=\square \\
x=\square
\end{array}
\]

Step 1 ：Given the function \(x^{2} y^{2}+x y=2\), we need to find the derivative to determine the slope of the tangent line at any point on the curve.

Step 2 ：The derivative of the function is \(2xy^2 + y\).

Step 3 ：We set the derivative equal to -1 to find the x-coordinates where the slope of the tangent line is -1. This gives us the equation \(2xy^2 + y + 1 = 0\).

Step 4 ：Solving this equation for x gives us \(x = \frac{-y - 1}{2y^2}\).

Step 5 ：We substitute this expression for x back into the original equation and solve for y, giving us \(y = -\frac{1}{3}\) and \(y = \frac{1}{3}\).

Step 6 ：Substituting these y-values back into the expression for x gives us the x-coordinates where the slope of the tangent line is -1: \(x = -3\) and \(x = -6\).

Step 7 ：\(\boxed{x=-3, x=-6}\) are the x-coordinates where the slope of the tangent line is -1.