1. Start with the given equation

$3\sin(\alpha+\beta)=2\sin(\alpha-\beta)$

Using angle-sum and angle-difference identities,

$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$ $\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$

So,

$3(\sin\alpha\cos\beta+\cos\alpha\sin\beta)=2(\sin\alpha\cos\beta-\cos\alpha\sin\beta)$

2. Expand and collect like terms

$3\sin\alpha\cos\beta+3\cos\alpha\sin\beta=2\sin\alpha\cos\beta-2\cos\alpha\sin\beta$

Bring all terms to one side:

$\sin\alpha\cos\beta+5\cos\alpha\sin\beta=0$

Thus,

$\sin\alpha\cos\beta=-5\cos\alpha\sin\beta$

3. Divide by $\cos\alpha\cos\beta$

Since $\alpha,\beta\in(0,\pi/2)$, we have $\cos\alpha>0$ and $\cos\beta>0$, so division is valid:

$\frac{\sin\alpha}{\cos\alpha}=-5\frac{\sin\beta}{\cos\beta}$

Hence,

$\tan\alpha=-5\tan\beta$

4. Compare with the given form

We are told

$\tan\alpha=k\tan\beta$

Therefore,

$k=-5$

5. Check the options

The correct option is:

$\boxed{-5}$

So option C is correct.