Answer
\[y=-9x+35\]
Work Step by Step
\[\begin{align}
& y=\frac{{{\left( {{x}^{2}}-1 \right)}^{2}}}{{{x}^{3}}-6x-1} \\
& \text{Differentiate} \\
& \frac{dy}{dx}=\frac{d}{dx}\left[ \frac{{{\left( {{x}^{2}}-1 \right)}^{2}}}{{{x}^{3}}-6x-1} \right] \\
& \frac{dy}{dx}=\frac{\left( {{x}^{3}}-6x-1 \right)\frac{d}{dx}\left[ {{\left( {{x}^{2}}-1 \right)}^{2}} \right]-{{\left( {{x}^{2}}-1 \right)}^{2}}\frac{d}{dx}\left[ {{x}^{3}}-6x-1 \right]}{{{\left( {{x}^{3}}-6x-1 \right)}^{2}}} \\
& \frac{dy}{dx}=\frac{\left( {{x}^{3}}-6x-1 \right)\left[ 2\left( {{x}^{2}}-1 \right)\left( 2x \right) \right]-{{\left( {{x}^{2}}-1 \right)}^{2}}\left( 3{{x}^{2}}-6 \right)}{{{\left( {{x}^{3}}-6x-1 \right)}^{2}}} \\
& \frac{dy}{dx}=\frac{4x\left( {{x}^{2}}-1 \right)\left( {{x}^{3}}-6x-1 \right)-\left( 3{{x}^{2}}-6 \right){{\left( {{x}^{2}}-1 \right)}^{2}}}{{{\left( {{x}^{3}}-6x-1 \right)}^{2}}} \\
& \text{Calculate the derivative at the point }\left( 3,8 \right) \\
& {{\left. \frac{dy}{dx} \right|}_{x=3}}=\frac{4\left( 3 \right)\left( {{3}^{2}}-1 \right)\left( {{3}^{3}}-6\left( 3 \right)-1 \right)-\left( 3{{\left( 3 \right)}^{2}}-6 \right){{\left( {{3}^{2}}-1 \right)}^{2}}}{{{\left( {{3}^{3}}-6\left( 3 \right)-1 \right)}^{2}}} \\
& \text{Simplifying} \\
& {{\left. \frac{dy}{dx} \right|}_{x=3}}=\frac{4\left( 3 \right)\left( 8 \right)\left( 8 \right)-\left( 27-6 \right){{\left( 8 \right)}^{2}}}{{{\left( 8 \right)}^{2}}} \\
& {{\left. \frac{dy}{dx} \right|}_{x=3}}=-9 \\
& m=-9 \\
& \text{Find the equation of the tangent line at the point }\left( 3,8 \right) \\
& y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\
& y-8=-9\left( x-3 \right) \\
& y-8=-9x+27 \\
& \text{ }y=-9x+35 \\
& \text{Graph} \\
\end{align}\]
