Answer
The asymptotes:
$y = \pm \sqrt 3 \left( {x + 1} \right) - 5$
Work Step by Step
Write
$3{x^2} + 6x - {y^2} - 10y = 1$
$3\left( {{x^2} + 2x} \right) - \left( {{y^2} + 10y} \right) = 1$
$3{\left( {x + 1} \right)^2} - 3 - {\left( {y + 5} \right)^2} + 25 = 1$
$3{\left( {x + 1} \right)^2} - {\left( {y + 5} \right)^2} + 22 = 1$
$3{\left( {x + 1} \right)^2} - {\left( {y + 5} \right)^2} = - 21$
Divide both sides by $-21$ gives
$\frac{1}{{21}}{\left( {y + 5} \right)^2} - \frac{1}{7}{\left( {x + 1} \right)^2} = 1$
${\left( {\frac{{y + 5}}{{\sqrt {21} }}} \right)^2} - {\left( {\frac{{x + 1}}{{\sqrt 7 }}} \right)^2} = 1$
This is the equation of a hyperbola where the foci are located on the $y$-axis, with $b = \sqrt {21} $, $a = \sqrt 7 $ and centered at $\left( { - 1, - 5} \right)$. If it were centered at the origin $\left( {0,0} \right)$, the asymptotes would be given by $x = \pm \frac{a}{b}y = \pm \frac{{\sqrt 7 }}{{\sqrt {21} }}y$. Thus, the asymptotes with center at $\left( { - 1,-5} \right)$ are
$x + 1 = \pm \frac{{\sqrt 7 }}{{\sqrt {21} }}\left( {y + 5} \right)$, ${\ \ }$ $x + 1 = \pm \frac{1}{{\sqrt 3 }}\left( {y + 5} \right)$
$y = \pm \sqrt 3 \left( {x + 1} \right) - 5$
