Solution
= Solution
Set $g(x)=\sqrt{8x}+\sin^2(2x)$. The <chain rule> gives $f'=-g'/g^2$, while
$$
g'(x)=\frac4{\sqrt{8x}}+4\sin(2x)\cos(2x)
=\frac4{\sqrt{8x}}+2\sin(4x).
$$
Thus
$$
\boxed{f'(x)=-\frac{4/\sqrt{8x}+2\sin(4x)}{\big[\sqrt{8x}+\sin^2(2x)\big]^2},\qquad x>0.}
$$
The real domain excludes zero because the original denominator vanishes there; for $x>0$ it is strictly positive.