Compute $a=f\circ g(10)$

First, $g(10)=\frac{10}{10-9}=10$ Hence, $a=f(g(10))=f(10)=10^2+9=109$

Compute $b=g\circ f(3)$

First, $f(3)=3^2+9=18$ Then, $b=g(18)=\frac{18}{18-9}=\frac{18}{9}=2$

So the ellipse is $\frac{x^2}{109}+\frac{y^2}{2}=1$

Identify major and minor axes

Since $109>2$, we compare with standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad a^2>b^2$ Thus, $a^2=109,\quad b^2=2$

Find eccentricity

For an ellipse, $e^2=1-\frac{b^2}{a^2}=1-\frac{2}{109}=\frac{107}{109}$

Find length of latus rectum

Length of latus rectum is $l=\frac{2b^2}{a}=\frac{2\cdot 2}{\sqrt{109}}=\frac{4}{\sqrt{109}}$ Therefore, $l^2=\frac{16}{109}$

Compute $8e^2+l^2$

$8e^2+l^2=8\cdot \frac{107}{109}+\frac{16}{109}$ $=\frac{856+16}{109}=\frac{872}{109}=8$

Check options

The value is $8$, so the correct option is: $\boxed{\text{C}}$