Answer
$$2$$
$$y=2 x+3 $$
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^{2}-2 h}-1}{h}\\
&=\lim _{h \rightarrow 0} \frac{1-1-h^{2}+2 h}{h\left(1+h^{2}-2 h\right)}\\
&=\lim _{h \rightarrow 0} \frac{-h^{2}+2 h}{h\left(1+h^{2}-2 h\right)}\\
&=\lim _{h \rightarrow 0} \frac{-h+2}{1+h^{2}-2 h}\\
&=2
\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&=2(x+1)\\
y&=1+2 x+2\\
y&=2 x+3
\end{align*}
