Answer
$$-3$$ $$y=-3 x+4$$
Work Step by Step
Since
\begin{aligned}
f^{\prime}(1)&=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}\\
&=\lim _{h \rightarrow 0} \frac{\frac{1}{1+h^{3}+3 h+3 h^{2}}-1}{h}\\
&=\lim _{h \rightarrow 0} \frac{1-1-h^{3}-3 h-3 h^{2}}{h\left(1+h^{3}+3 h+3 h^{2}\right)}\\
&=\lim _{h \rightarrow 0} \frac{-h^{3}-3 h-3 h^{2}}{h\left(1+h^{3}+3 h+3 h^{2}\right)}\\
&=\lim _{h \rightarrow 0} \frac{-h^{2}-3-3 h}{1+h^{3}+3 h+3 h^{2}}\\
&=-3
\end{aligned}
Thus at $a=1,$ the equation of the tangent line is $y-f(1)=f^{\prime}(1)(x-1)$
\begin{align*}
y-1&=-3(x-1)\\
y&=1-3 x+3\\
y&=-3 x+4
\end{align*}
