Given:

$x^2+y^2-6x+4y+12=0 ....\left(1\right)$

$\Rightarrow {\left(x-3\right)}^2+{\left(y+2\right)}^2=1$

It represents circle having centre $\left(3,-2\right)$ and radius $1$ unit.

Let the slope of tangent be $m$, then equation of tangent is

$y+3=m\left(x-1\right)$

$\Rightarrow mx-y-m-3=0 ....\left(2\right)$

Perpendicular distance from centre of the circle to the tangent must be equal to the radius of the circle, therefore

$\left|\frac{3m+2-m-3}{\sqrt{1+m^2}}\right|=1$

$\Rightarrow \frac{2m-1}{\sqrt{1+m^2}}=\pm 1$

$\Rightarrow {\left(2m-1\right)}^2=1+m^2$

$\Rightarrow 3m^2-4m=0$

$\Rightarrow m=0, \frac{4}{3}$

So,

$m_1=0, m_2=\frac{4}{3}$

$\therefore 9\left(m_1^2+m_2^2\right)=9\times \frac{16}{9}=16$