1. **State the problem:** Differentiate the given implicit function $$e^{x+y} - 3x + 2 = y$$ with respect to $x$. Then evaluate the derivative at the point $\left( \ln 2, 0 \right)$. 2. **Rewrite the equation:** $$e^{x+y} - 3x + 2 = y$$ 3. **Differentiate implicitly:** Treat $y$ as a function of $x$, so use the chain rule for terms involving $y$. $$\frac{d}{dx}\left(e^{x+y}\right) - \frac{d}{dx}(3x) + \frac{d}{dx}(2) = \frac{dy}{dx}$$ 4. **Calculate each derivative:** - $$\frac{d}{dx}\left(e^{x+y}\right) = e^{x+y} \cdot \frac{d}{dx}(x+y) = e^{x+y} (1 + y')$$ - $$\frac{d}{dx}(3x) = 3$$ - $$\frac{d}{dx}(2) = 0$$ - $$\frac{d}{dx}(y) = y'$$ 5. **Write the differentiated equation:** $$e^{x+y} (1 + y') - 3 = y'$$ 6. **Rearrange terms to isolate $y'$:** $$e^{x+y} + e^{x+y} y' - 3 = y'$$ 7. **Group $y'$ terms:** $$e^{x+y} y' - y' = 3 - e^{x+y}$$ 8. **Factor out $y'$:** $$y' (e^{x+y} - 1) = 3 - e^{x+y}$$ 9. **Solve for $y'$:** $$y' = \frac{3 - e^{x+y}}{e^{x+y} - 1}$$ 10. **Evaluate $y'$ at $\left(\ln 2, 0\right)$:** - Compute $e^{x+y} = e^{\ln 2 + 0} = 2$ - Substitute into formula: $$y' = \frac{3 - 2}{2 - 1} = \frac{1}{1} = 1$$ **Final answer:** $$\frac{dy}{dx} \bigg|_{(\ln 2, 0)} = 1$$