By making the substitution u-(sin)^(-1) (x/3), simplify: x^2/√(9-x^2 )
$\displaystyle \displaystyle u=\sin^{-1} \frac{x}{3}$
$\displaystyle \displaystyle \sin u= \frac{x}{3}$
$\displaystyle \displaystyle x = 3\sin u$
So the expression becomes
$\displaystyle \displaystyle \frac{x^2}{\sqrt{9-x^2}} = \frac{9\sin^2 u}{\sqrt{9-(3\sin u)^2}} = \dots$
Can you finish it from here?
Do you need to integrate this function? If so, consider $\displaystyle \displaystyle x = 3\sin u \implies dx = 3\cos u~du$
[3 sin(u)]^2 / sqrt(9 - [3 sin(u)]^2)
[ 9 sin^2(u) ] / sqrt(9 - 9 sin^2(u))
[ 9 sin^2(u) ] / sqrt(9{1 - sin^2(u)})
[ 9 sin^2(u) ] / sqrt(9cos^2(u))
[ 9 sin^2(u) ] / [3 cos(u)]
[ 3 sin^2(u) ] / [ cos(u)]
3 tan(u) sin(u)
YEa?