Step 1:

Let $P(x,y)$ be the point which moves such that sum of its distances from $F(3,0)$ and $F'(-3,0)$ is $9$

The curve traced by $P$ is an ellipse length of whose major axes $(2a=9)$ and whose central is at the midpoint of $FF'$ at $(0,0)$,where $F,F'$ are the foci.

Step 2:

The major axis is the $x$-axis since the foci lie on it.

Now $CF=ae=3$

$a=\large\frac{9}{2}$

Therefore $e=\large\frac{3}{a}=\frac{3\times 2}{9}=\frac{2}{3}$

Step 3:

$b=a\sqrt{1-e^2}$

$\Rightarrow \large\frac{9}{2}\sqrt{1-\large\frac{4}{9}}=\frac{3\sqrt 5}{2}$

$b^2=\large\frac{45}{4}$

The equation is $\large\frac{x^2}{81/4}+\frac{y^2}{45/4}$$=1$