To find the value of $$k$$, the given conditions are:

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

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

For the first equation, using the sum and difference formulas for sine, we can rewrite the equation as:

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

Simplifying this, we get:

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

Rearranging the terms, we obtain:

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

Dividing both sides by $$\sin \alpha \cos \beta$$, we get:

$$\frac{5 \sin \beta \cos \alpha}{\sin \alpha \cos \beta} = -1$$

Which simplifies to:

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

So, taking the reciprocal, we have:

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

Therefore, by comparing this equation with the given $$\tan \alpha = k \tan \beta$$, we find that $$k = -5$$.

Thus, the value of $$k$$ is $$-5$$.