After applying implicit differentiation to the given equation and solving for dy/dx, the slope expression in terms of x and y is obtained. By substituting x=0 into the original equation, y is found to be -e^2. The equation of the tangent line at the point (0, -e^2) with a slope of 1 is y = x - e^2, which simplifies to y = x - 7.3891 after rounding.
To find the slope of the tangent line and the equation of the tangent line at a specific point using implicit differentiation, let's differentiate the given equation ln(x - y) + 2 = 3x² with respect to x. Applying implicit differentiation, we have:
\(rac{1}{x - y}(1 - \frac{dy}{dx}) = 6x\). Now, we can solve for \(\frac{dy}{dx}\) to find the expression of the slope of the tangent line in terms of x and y:
\(\frac{dy}{dx} = 1 - 6x(x - y)\).
Substituting x = 0 into the original equation to find y, we get ln(-y) + 2 = 0, which simplifies to y = -e^2. Using x = 0 and y = -e^2 in the slope expression, we find that the slope at the point (0, -e^2) is \(1\).
The equation of the tangent line is then determined using the point-slope form y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point through which the line passes. Therefore, the equation of the tangent line at (0, -e^2) is:
y + e^2 = 1(x - 0) or y = x - e^2.
After rounding the coefficient of e^2 to four decimal places, we get the final answer:
y = x - 7.3891.