1. **State the problem:** We need to find the slope of the tangent line to the curve defined by the implicit equation $$x^2 + 4xy - 3y^2 = 7$$ at the point $(2,1)$. 2. **Formula and rules:** The slope of the tangent line to an implicitly defined curve is given by $$\frac{dy}{dx}$$, which can be found using implicit differentiation. 3. **Differentiate both sides with respect to $x$:** $$\frac{d}{dx}(x^2) + \frac{d}{dx}(4xy) - \frac{d}{dx}(3y^2) = \frac{d}{dx}(7)$$ 4. **Apply differentiation rules:** - $$\frac{d}{dx}(x^2) = 2x$$ - For $$4xy$$, use the product rule: $$\frac{d}{dx}(4xy) = 4\left(y + x\frac{dy}{dx}\right) = 4y + 4x\frac{dy}{dx}$$ - For $$3y^2$$, use the chain rule: $$\frac{d}{dx}(3y^2) = 6y\frac{dy}{dx}$$ - The derivative of a constant is zero. 5. **Substitute these into the differentiated equation:** $$2x + 4y + 4x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0$$ 6. **Group terms with $$\frac{dy}{dx}$$ on one side:** $$4x\frac{dy}{dx} - 6y\frac{dy}{dx} = -2x - 4y$$ 7. **Factor out $$\frac{dy}{dx}$$:** $$\frac{dy}{dx}(4x - 6y) = -2x - 4y$$ 8. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = \frac{-2x - 4y}{4x - 6y}$$ 9. **Evaluate at the point $(2,1)$:** $$\frac{dy}{dx}\bigg|_{(2,1)} = \frac{-2(2) - 4(1)}{4(2) - 6(1)} = \frac{-4 - 4}{8 - 6} = \frac{-8}{2} = -4$$ **Final answer:** The slope of the tangent line at the point $(2,1)$ is $$\boxed{-4}$$.