1. **Problem a:** Find the equation of the tangent line to the curve defined by $$x^2 y + y^2 x = 6$$ at the point $$P(1,2)$$. 2. **Formula and rules:** To find the tangent line, we need the derivative $$\frac{dy}{dx}$$ using implicit differentiation because $$y$$ is implicit in the equation. 3. Differentiate both sides with respect to $$x$$: $$\frac{d}{dx}(x^2 y) + \frac{d}{dx}(y^2 x) = \frac{d}{dx}(6)$$ 4. Use the product rule: $$\frac{d}{dx}(x^2 y) = 2x y + x^2 \frac{dy}{dx}$$ $$\frac{d}{dx}(y^2 x) = y^2 + 2y x \frac{dy}{dx}$$ 5. Substitute back: $$2x y + x^2 \frac{dy}{dx} + y^2 + 2y x \frac{dy}{dx} = 0$$ 6. Group terms with $$\frac{dy}{dx}$$: $$x^2 \frac{dy}{dx} + 2y x \frac{dy}{dx} = -2x y - y^2$$ 7. Factor $$\frac{dy}{dx}$$: $$\frac{dy}{dx}(x^2 + 2y x) = -2x y - y^2$$ 8. Solve for $$\frac{dy}{dx}$$: $$\frac{dy}{dx} = \frac{-2x y - y^2}{x^2 + 2y x}$$ 9. Evaluate at $$P(1,2)$$: $$\frac{dy}{dx} = \frac{-2(1)(2) - (2)^2}{(1)^2 + 2(2)(1)} = \frac{-4 - 4}{1 + 4} = \frac{-8}{5}$$ 10. Equation of tangent line at $$P(1,2)$$: $$y - 2 = \frac{-8}{5}(x - 1)$$ 11. Simplify: $$y = 2 - \frac{8}{5}(x - 1) = 2 - \frac{8}{5}x + \frac{8}{5} = \frac{18}{5} - \frac{8}{5}x$$ --- **Final answer:** $$y = \frac{18}{5} - \frac{8}{5}x$$