1. First, restate the problem: Differentiate the equation $$e^{x+y} - 3xy - 2 = y$$ with respect to $$x$$. 2. We will use implicit differentiation because $$y$$ is a function of $$x$$. 3. Differentiate each term: - For $$e^{x+y}$$, use the chain rule: $$\frac{d}{dx} e^{x+y} = e^{x+y} \cdot \frac{d}{dx} (x+y) = e^{x+y} (1 + \frac{dy}{dx})$$. - For $$-3xy$$, use the product rule: $$\frac{d}{dx} (-3xy) = -3 \left(y + x \frac{dy}{dx}\right) = -3y - 3x \frac{dy}{dx}$$. - For $$-2$$, derivative is $$0$$ because it is constant. - For right side $$y$$, derivative is $$\frac{dy}{dx}$$. 4. Putting it all together: $$e^{x+y} (1 + \frac{dy}{dx}) - 3y - 3x \frac{dy}{dx} = \frac{dy}{dx}$$. 5. Expand: $$e^{x+y} + e^{x+y} \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = \frac{dy}{dx}$$. 6. Group all $$\frac{dy}{dx}$$ terms on the left and others on the right: $$e^{x+y} \frac{dy}{dx} - 3x \frac{dy}{dx} - \frac{dy}{dx} = -e^{x+y} + 3y$$. 7. Factor out $$\frac{dy}{dx}$$: $$\frac{dy}{dx} (e^{x+y} - 3x - 1) = -e^{x+y} + 3y$$. 8. Solve for $$\frac{dy}{dx}$$: $$\frac{dy}{dx} = \frac{-e^{x+y} + 3y}{e^{x+y} - 3x - 1}$$. Final answer: $$\boxed{\frac{dy}{dx} = \frac{3y - e^{x+y}}{e^{x+y} - 3x - 1}}$$