Answer
$(-3,0)$
Work Step by Step
Given $$y = [\ln(x+4)]^2$$
Since
\begin{align*}
y'&= 2[\ln(x+4)](x+4)'\\
&=\frac{2\ln(x+4)}{x+4}
\end{align*}
The tangent is horizontal when $y'=0$
\begin{align*}
y'&=0\\
\frac{2\ln(x+4)}{x+4}&=0\\
\ln(x+4)&=0\\
x+4&=1\\
x&=-3
\end{align*}
and $y=0$, then at $(-3,0)$ the tangent is horizontal