To solve this problem, let's use implicit differentiation step by step.

The equation given is: $\ln(x^2 - 8y) = x - y - 2$ with $y(3) = 1$. We are asked to find $y'(3)$ and then use it to find the equation of the tangent line at the point $(3, 1)$.

Step 1: Implicit Differentiation

Differentiating both sides of the equation with respect to $x$:

1. Differentiate the left side: $\frac{d}{dx} \left( \ln(x^2 - 8y) \right) = \frac{1}{x^2 - 8y} \cdot (2x - 8y')$

2. Differentiate the right side: $\frac{d}{dx} (x - y - 2) = 1 - y'$

So, putting it all together: $\frac{2x - 8y'}{x^2 - 8y} = 1 - y'$

Step 2: Solve for $y'$

Rearrange to solve for $y'$:

$2x - 8y' = (x^2 - 8y)(1 - y')$ Expanding and isolating $y'$:

$2x - 8y' = x^2 - 8y - x^2 y' + 8yy'$

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