Absolute Minimum will be at $x = 3$ and $x = -3$.
Absolute minimum of $f\left ( x \right )$ is the minimum possible value that $f\left ( x \right )$ can ever attain.
Since a square root never spits out a $-ve$ value, the minimum value that $f\left ( x \right )$ can attain is $0$.
Now to make $f\left ( x \right ) = 0$, $\left ( 36 -4x^2 \right )$ must be equal to $0$.
On solving $\left ( 36 -4x^2 \right )= 0$, we get $x = 3$ & $x = -3$.
So at $x = 3$ & $x = -3$ the function $f\left ( x \right )$ will be at its absolute minimum.
that is, $f\left ( 3 \right ) = f\left ( -3 \right ) = 0$.
Also the domain of $f\left ( x \right )$ is $\left [ -3, 3 \right ]$ & the range is $\left [ 0, 6 \right ]$