Answer
$\frac{9}{\sqrt{9-x^2}(9-x^2)}$
Work Step by Step
First, re-write to make it in radical form:
$\frac{{\sqrt{9-x^2}}+\frac{x^2}{\sqrt{9-x^2}}}{9-x^2}$
Then, make both denominators of the top fraction the same:
$\frac{\frac{\sqrt{9-x^2}\cdot\sqrt{9-x^2}}{\sqrt{9-x^2}}+\frac{x^2}{\sqrt{9-x^2}}}{9-x^2}$
Because of the law of radicals $\sqrt[n] a \cdot \sqrt[n] b=\sqrt[n] {ab}$:
$\frac{\frac{\sqrt{(9-x^2)^2}}{\sqrt{9-x^2}}+\frac{x^2}{\sqrt{9-x^2}}}{9-x^2}$
Simplify further:
$\frac{\frac{9-x^2+x^2}{\sqrt{9-x^2}}}{9-x^2}=\frac{9}{\sqrt{9-x^2}}\cdot\frac{1}{9-x^2}=\frac{9}{\sqrt{9-x^2}(9-x^2)}$
