Answer: \(\dfrac{dy}{dx}=2\sqrt3\) at \(x=\dfrac{\pi}{3}\); approximate change \(2\sqrt3h\); \(\dfrac{dy}{dt}=6\sqrt3\).

Since

\(y=\frac{\tan3x}{\sin x},\)

use the quotient rule. This gives

\(\frac{dy}{dx} =\frac{\sin x(3\operatorname{sec}^2 3x)-\tan3x\cos x}{\sin^2x}.\)

When \(x=\frac{\pi}{3}\),

\(\sin x=\frac{\sqrt3}{2},\qquad \tan3x=\tan\pi=0,\qquad \operatorname{sec}^2 3x=\operatorname{sec}^2\pi=1.\)

Therefore

\(\frac{dy}{dx} =\frac{\frac{\sqrt3}{2}\cdot3-0}{\left(\frac{\sqrt3}{2}\right)^2} =\frac{3\sqrt3/2}{3/4} =2\sqrt3.\)

For a small increase \(h\), the approximate change in \(y\) is

\(\Delta y\approx \frac{dy}{dx}h=2\sqrt3h.\)

For the rate of change,

\(\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}.\)

Since \(\frac{dx}{dt}=3\),

\(\frac{dy}{dt}=2\sqrt3\times3=6\sqrt3.\)