Answer
$$-1$$
$$y= -x-1 $$
Work Step by Step
Since
$f(x)=\dfrac{1}{x+3} .$ Then
\begin{align*}
f^{\prime}(-2)&=\lim _{h \rightarrow 0} \frac{f(-2+h)-f(-2)}{h}\\
&=\lim _{h \rightarrow 0} \frac{\frac{1}{-2+h+3}-1}{h}\\
&=\lim _{h \rightarrow 0} \frac{\frac{1}{1+h}-1}{h}\\
&=\lim _{h \rightarrow 0} \frac{-h}{h(1+h)}\\
&=\lim _{h \rightarrow 0} \frac{-1}{1+h}=-1\end{align*}
The tangent line at $a=-2$ is
$$
y=f^{\prime}(-2)(x+2)+f(-2)=-1(x+2)+1=-x-1
$$
