Answer
Decreasing on: $(-4, -2\sqrt{2}), (2\sqrt{2},4)$
Increasing on: $(-2\sqrt{2},2\sqrt{2})$
Work Step by Step
$y=x\sqrt{16-x^{2}}$ , defined on $[-4, 4]$
$y^{\prime}=\displaystyle \frac{-2(x^{2}-8)}{\sqrt{16-x^{2}}}$, defined on $(-4, 4)$
$=\displaystyle \frac{-2(x-2\sqrt{2})(x+2\sqrt{2})}{\sqrt{16-x^{2}}}$
Critical numbers: $x=\pm 2\sqrt{2}$
$\left[\begin{array}{llll}
Interval & (-4,-2\sqrt{2}) & (-2\sqrt{2},2\sqrt{2}) & (2\sqrt{2},\infty)\\
\text{test point} & -3 & 0 & 3\\
f^{\prime}(\text{test point}) & -0.7559 & 4 & -0.7559\\
\text{sign} & - & + & -\\
& \searrow & \nearrow & \searrow
\end{array}\right]$
Decreasing on: $(-4, -2\sqrt{2}), (2\sqrt{2},4)$
Increasing on: $(-2\sqrt{2},2\sqrt{2})$
