Finding the second directrix of a hyperbola

One focus and the nearest directrix to this focus are $$(0,-5)$$ and $$d_1: x-3y-4=0$$, respectively. Also, the eccentricity is $$\sqrt{10}$$. Find the second directrix of the hyperbola.

I assumed the second directrix is $$d_2: x-3y+c=0$$ as directrices are parallel. And then, I got two possible values for $$c$$, namely, $$-\cfrac{14}{9}; -\cfrac{256}{9}$$.

How to decide which to choose?

One of your roots corresponds to a hyperbola, which is what you want. The second root corresponds to an ellipse. The ellipse in the second root has an eccentricity equal to $$1$$ divided by the eccentricity of the hyperbola you want. A key difference between the two conic sections is that a hyperbola has its directrices between the foci, while for an ellipse it's the other way around.
To orient yourself, find the value of $$c$$ such that $$x-3y+c=0$$ at the given focus. This, of course, is $$-15$$. The given directrix has $$c=-4$$, which is greater. So now, if you try the root $$c=-256/9<-15$$ for the other directrix, you have the focus between the directrices which corresponds to that ellipse. To get the hyperbola you must select $$c>-4$$ for the other directrix, putting the given directrix ($$c=-4$$) between the focus ($$c=-15$$) and the directrix you're trying to find. Therefore select $$c=-14/9$$.