Answer: \(\dfrac{dy}{dx}=\dfrac{\cos x-3\sin x}{\sin x+3\cos x}\); \(x=\dfrac{\pi}{4}\).

Let

\(u=\sin x+3\cos x.\)

Then

\(y=\ln u\)

and

\(\frac{du}{dx}=\cos x-3\sin x.\)

So

\(\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx} =\frac{\cos x-3\sin x}{\sin x+3\cos x}.\)

Now solve

\(\frac{\cos x-3\sin x}{\sin x+3\cos x}=-\frac12.\)

Multiplying by \(2(\sin x+3\cos x)\),

\(2\cos x-6\sin x=-\sin x-3\cos x.\)

Hence

\(5\cos x=5\sin x.\)

So

\(\tan x=1.\)

Since \(0\lt x\lt\frac{\pi}{2}\),

\(x=\frac{\pi}{4}.\)