Answer
$y=3x-9$
Work Step by Step
$y=\ln \left(x^{2}-3 x+1\right)\\
y^{\prime}=\frac{1}{x^{2}-3 x+1} \times 2 x-3\\
y^{\prime}=\frac{2 x-3}{x^{2}-3 x+1}$
Gradient at point $(3,0)$
$y^{\prime}=\frac{2(3)-3}{(3)^{2}-3(3)+1}\\
y^{\prime}=3$
Tangent line at point $(3,0)$
$y-0 =3(x-3)\\
y=3x-9$
