Why not the domain is just $ ( - \infty , -3) $ of $ \sqrt{2-x} - \frac{1}{9-x^2} $ Consider

$$ f(x) = \sqrt{2-x} - \frac{1}{9-x^2} $$

Now for the radical to be definable, 


  $$ 2-x \ge 0 $$


And for the fraction to be definable,


  $$ 9-x^2&gt; 0 $$


SO , the number line of solution set looks like:


Where red line is for radical's definability &amp; orange for fraction's.

Now,

The result from above stuff is 


  $$ (- \infty , -3) $$ 


EDIT:
I got that the second inequality's solution set was wrong. 

Its 


  $$ (-3,3)$$ 


So final solution plot:



Which is  $ \color{red}{wrong} $ 

Why?
The fraction is defined if and only if $9-x^2\ne 0\Leftrightarrow x\ne\pm3.$

However, you have done the square root part correctly: $\sqrt{2-x}$ is defined if and only if $x\le 2$.

Because of this, we can conclude that $f(x)$ is defined if and only if $x\le 2$ and $x\ne -3$.
In one line:

$ x \le 2$ AND $(x \not = -3$ OR $x \not =3)$.

Hence $x \le 2$ AND $x \not = -3.$

$D= (-\infty, -3)\cup (-3,2]$.