Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 33

The equation of the tangent line to the curve y = ln (x^2 - 3x + 1) , at the point (3,0) is the following, such that:
f(x) - f(x_0) = f'(x_0)(x - x_0)
You need to put f(x) = y, f(x_0) = 0, x_0 = 3
You need to evaluate f'(3), hence, you need to find the derivative of the function, using the product rule, such that:
f'(x) = (ln (x^2 - 3x + 1))'*(x^2 - 3x + 1)'
f'(x) = (1/(x^2 - 3x + 1))*(2x - 3)
Replacing 3 for x, yields:
f'(3) = (1/(3^2 - 3*3 + 1))*(2*3 - 3)
f'(3) = 3/1 => f'(3) = 3
Replacing the values in the equation of tangent line, yields:
y - 0 = 3*(x - 3)
y = 3*(x - 3)
Hence, evaluating the equation of the tangent line to the given curve, at the point (3,0), yields y = 3*(x - 3).