1. **State the problem:** Find the equation of the tangent line to the curve defined by $$ (x^2 + x^2)^3 = 8x^2y^2 $$ at the point $(-1, 1)$. 2. **Simplify the equation:** Note that $x^2 + x^2 = 2x^2$, so the equation becomes $$ (2x^2)^3 = 8x^2y^2 $$ which simplifies to $$ 8x^6 = 8x^2y^2 $$ 3. **Divide both sides by 8:** $$ x^6 = x^2 y^2 $$ 4. **Rewrite the equation:** $$ x^6 - x^2 y^2 = 0 $$ 5. **Implicit differentiation:** Differentiate both sides with respect to $x$. Using the product and chain rules, $$ \frac{d}{dx}(x^6) - \frac{d}{dx}(x^2 y^2) = 0 $$ 6. **Calculate derivatives:** $$ 6x^5 - \left( 2x y^2 + x^2 \cdot 2y \frac{dy}{dx} \right) = 0 $$ 7. **Simplify:** $$ 6x^5 - 2x y^2 - 2x^2 y \frac{dy}{dx} = 0 $$ 8. **Solve for $\frac{dy}{dx}$:** $$ 2x^2 y \frac{dy}{dx} = 6x^5 - 2x y^2 $$ $$ \frac{dy}{dx} = \frac{6x^5 - 2x y^2}{2x^2 y} $$ 9. **Evaluate at the point $(-1, 1)$:** $$ \frac{dy}{dx} = \frac{6(-1)^5 - 2(-1)(1)^2}{2(-1)^2 (1)} = \frac{6(-1) - 2(-1)}{2(1)(1)} = \frac{-6 + 2}{2} = \frac{-4}{2} = -2 $$ 10. **Find the tangent line equation:** Using point-slope form $$ y - y_1 = m(x - x_1) $$ with $m = -2$ and point $(-1, 1)$, $$ y - 1 = -2(x + 1) $$ $$ y - 1 = -2x - 2 $$ $$ y = -2x - 1 $$ **Final answer:** The equation of the tangent line at $(-1, 1)$ is $$ y = -2x - 1 $$