1. **State the problem:** Find the slope of the tangent line to the curve defined by the implicit equation $$xy^2 - 2xy + x^2y = 0$$ at the point $$(x,y) = (1,1)$$. 2. **Recall the formula:** To find the slope of the tangent line to an implicitly defined curve, we use implicit differentiation to find $$\frac{dy}{dx}$$. 3. **Differentiate both sides with respect to $$x$$:** $$\frac{d}{dx}(xy^2) - \frac{d}{dx}(2xy) + \frac{d}{dx}(x^2y) = \frac{d}{dx}(0)$$ 4. **Apply the product rule to each term:** - For $$xy^2$$: $$\frac{d}{dx}(x y^2) = y^2 + x \cdot 2y \frac{dy}{dx} = y^2 + 2xy \frac{dy}{dx}$$ - For $$2xy$$: $$\frac{d}{dx}(2xy) = 2\left(y + x \frac{dy}{dx}\right) = 2y + 2x \frac{dy}{dx}$$ - For $$x^2 y$$: $$\frac{d}{dx}(x^2 y) = 2x y + x^2 \frac{dy}{dx}$$ 5. **Substitute these into the differentiated equation:** $$y^2 + 2xy \frac{dy}{dx} - (2y + 2x \frac{dy}{dx}) + (2x y + x^2 \frac{dy}{dx}) = 0$$ 6. **Group terms with $$\frac{dy}{dx}$$ and without:** $$\left(2xy - 2x + x^2\right) \frac{dy}{dx} + (y^2 - 2y + 2xy) = 0$$ 7. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = - \frac{y^2 - 2y + 2xy}{2xy - 2x + x^2}$$ 8. **Evaluate at the point $$(1,1)$$:** - Numerator: $$1^2 - 2(1) + 2(1)(1) = 1 - 2 + 2 = 1$$ - Denominator: $$2(1)(1) - 2(1) + 1^2 = 2 - 2 + 1 = 1$$ 9. **Calculate the slope:** $$\frac{dy}{dx}\bigg|_{(1,1)} = - \frac{1}{1} = -1$$ **Final answer:** The slope of the tangent line at $$(1,1)$$ is $$-1$$.