#### Answer

The domain is
$$\mathcal{D}=\left\{(x,y)\left|\frac{x^2}{3^2}+\frac{y^2}{1^2}<1\right.\right\},$$
and it is sketched in the figure bellow

#### Work Step by Step

The argument of the logarithm has to be greater than zero i.e. we need that $9-x^2-9y^2>0$. Rearranging terms this gives
$$x^2+9y^2<9.$$
Dividing by $9$ we get
$$\frac{x^2}{9}+y^2<1.$$
Rewrite previous inequality as
$$\frac{x^2}{3^2}+\frac{y^2}{1^2}<1.$$
Knowing that
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}<1$$ is equation of a centered ellipse with major semiaxis equal to $a$ and minor semiaxis equal to $b$, we see that the domain is the region bounded by the ellipse with major semiaxis equal to $3$ and minor semiaxis equal to $1$ (the region does not include points from the ellipse because we have strictly less sign in the inequality).
We can write for the domain
$$\mathcal{D}=\left\{(x,y)\left|\frac{x^2}{3^2}+\frac{y^2}{1^2}<1\right.\right\},$$
and it is shown in the figure below