Answer
Solution set = $\{-1\}$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{lll}
x+3 \gt 0, & and & x+4 \gt 0\\
x \gt -3 & & x \gt -4
\end{array}\right]$
$ x \gt -3 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log_{6}(x+3)+\log_{6}(x+4)=1$
LHS: apply the product rule; RHS write $1$ as $\log_{6}6$
$\log_{6}[(x+3)(x+4)]=\log_{6}6\qquad $
(if $\log_{b}M=\log_{b}N,$ then M=N)
$(x+3)(x+4)=6$
$ x^{2}+7x+12=6$
$ x^{2}+7x+6=0$
Factor, by finding two factors of $6$ whose sum is $7$
$(x+6)(x+1)=0$
$ x=-6$ or $ x=-1$
The solution $ x=-6$ does not satisfy (*), so the
solution set = $\{-1\}$
