45 баллов! Найти область определения

$y= \frac{\sqrt{9-x^2}}{x+1} \\ \\ \left \{ {{ \sqrt{9-x^2} \geq 0\\} \atop {x+1 \neq 0}} \right. \\ \\ \left \{ {{9-x^2 \geq 0\\} \atop {x \neq -1}} \right. \\ \\ \left \{ {{-x^2 \geq -9\\} \atop {x \neq -1}} \right. \\ \\ \left \{ {{x^2 \leq 9\\} \atop {x \neq -1}} \right. \\ \\ \left \{ {{x \leq 3;x \geq-3\\} \atop {x \neq -1}} \right. \\ \\$

x ∈ [ -3; -1 ) U ( -1; 3 ]