There are 3 possible ways they can meet: Sevilla vs Barcelona, Sevilla vs Real Madrid, Real Madrid vs Barcelona. These are disjoint, and the chance of any particular draw happening is $1/7$ (since any one team can play any of the other $7$, randomly. So the probability of two spanish teams meeting is $3/7$, and the probability of avoiding each other is $4/7$.
An alternative method is to consider this: the first spanish team is picked. There are $2/7$ ways for them to match one of the other two spanish teams. Suppose they don't. Then from the remaining $6$ teams, the chance of the two spanish teams matching is $1/5$, by the fact that any pairing is equally likely from the perspective of one of these remaining teams. So the probability of matching up is $$\frac27+\frac57\cdot\frac15=\frac37$$ as before.