I am trying to solve the inequality $\log_{\log{\sqrt{9-x^2}}} x^2 <0$ I am trying to solve  the inequality 
$$\log_{\log{\sqrt{9-x^2}}}   x^2 <0.$$
I got $\mathrm{S.S}=(-\sqrt8 ,-1)\cup( 1,\sqrt8)$, but a friend  got  $\mathrm{S.S}=(-1,1)- \{0\}$.
Please, what is true?

$$\log_{\log_c(\sqrt{9-x^2})}x^2=\frac{\log_b x^2}{\log_b(\log_c\sqrt{9-x^2})}<0 $$
Without any loss of generality, we can take base $b>1$
(i)If  $\log_b x^2<0 \iff x^2<1$
then we need 
$\log_b(\log_c\sqrt{9-x^2})>0\implies \log_c\sqrt{9-x^2}>1\implies x^2<9-c^2$
$\implies x^2<min(1,9-c^2)$
Here observe that for real $x, min(1,9-c^2)>x^2>0\implies 9-c^2>0\implies c^2<9$ else there will be no solution.
(ii)If  $\log_b x^2>0 \iff x^2>1$
then we need $\log_b(\log_c\sqrt{9-x^2})<0\implies\log_c\sqrt{9-x^2}<1\implies x^2>9-c^2$
$\implies x^2>max(1,9-c^2)$
