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PREVIOUS ANSWERS SCALC9 2.6.061.
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Find all points on the curve $x^{2} y^{2}+x y=2$ where the slope of the tangent line is -1 . (Enter your answers as a commaseparated list of ordered pairs.)
\[
(x, y)=\sqrt{\frac{2}{3}},-\sqrt{\frac{2}{3}}
\]
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Step 1 ：The given equation is \(x^{2} y^{2}+x y=2\). We need to find the points where the slope of the tangent line is -1.

Step 2 ：First, we need to find the derivative of the given equation. We can use the product rule for differentiation, which states that the derivative of two functions multiplied together is the first function times the derivative of the second function plus the second function times the derivative of the first function.

Step 3 ：Differentiating \(x^{2} y^{2}+x y=2\) with respect to \(x\), we get \(2x y^{2} + x^{2} 2y y' + y + x y' = 0\).

Step 4 ：We can rearrange this equation to solve for \(y'\), the derivative of \(y\) with respect to \(x\). This gives us \(y' = -\frac{2x y^{2} + y}{2x^{2} y + x}\).

Step 5 ：We set \(y' = -1\) and solve for \(x\) and \(y\). This gives us the equation \(2x y^{2} + y = 2x^{2} y + x\).

Step 6 ：Solving this equation, we get \(x = \sqrt{\frac{2}{3}}, -\sqrt{\frac{2}{3}}\).

Step 7 ：Substituting these values of \(x\) back into the original equation, we can solve for \(y\).

Step 8 ：For \(x = \sqrt{\frac{2}{3}}\), we get \(y = \sqrt{\frac{2}{3}}, -\sqrt{\frac{2}{3}}\).

Step 9 ：For \(x = -\sqrt{\frac{2}{3}}\), we get \(y = \sqrt{\frac{2}{3}}, -\sqrt{\frac{2}{3}}\).

Step 10 ：Therefore, the points on the curve where the slope of the tangent line is -1 are \((\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}})\), \((\sqrt{\frac{2}{3}}, -\sqrt{\frac{2}{3}})\), \((-\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}})\), and \((-\sqrt{\frac{2}{3}}, -\sqrt{\frac{2}{3}})\).

Step 11 ：\boxed{\text{The points are } (\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}), (\sqrt{\frac{2}{3}}, -\sqrt{\frac{2}{3}}), (-\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}), (-\sqrt{\frac{2}{3}}, -\sqrt{\frac{2}{3}})}.