Given that $2g(x)+x\sin\!\left(g(x)\right) = 18x^{2}+47x+5$ and $g \bigl( -{\frac{1}{9}} \bigr) = 0$, find $g' \bigl( -{\frac{1}{9}} \bigr)$.

$g' \bigl( -{\frac{1}{9}} \bigr) =$