Let $\frac{y}{x}=m^{\text{'}}$. So, on putting $y=m^{\prime }x$ in the given circle,
We get ${\left(x-3\right)}^2+{\left(m^{\prime }x-3\right)}^2=6$
$\Rightarrow x^2\left(1+m^{{\text{'}}^{}2}\right)-6x\left(1+m^{\text{'}}\right)+12=0$
Putting discriminant $=0$, we get,
$m^{\text{'}}=3\underset{\_}{+}2\sqrt{2}$
So, m = $3-2\sqrt{2}$ and $M=3+2\sqrt{2}\Rightarrow mM=1$
$\Rightarrow \frac{1}{M}+\frac{1}{m}=m+M$
$=3-2\sqrt{2}+3+2\sqrt{2}$
$=6$