You have listed two properties for something to be a PRF, efficiency and pseudorandomness.
The function $F'$ is clearly efficient as $F$ is efficient and $F'$ calls $F$ in just about every case. The one case where it doesn't is a very easily computable case too.
So, that leaves pseudorandomness. What that property is basically saying is imagine you are given a black box. You can feed it inputs (the x and k values) and see the outputs. If there exists a probabilistic polynomial time (PPT) algorithm that can, given the outputs, decide whether or not the black box implements $F'$ vs a random function, you do not have a PRF. On the other hand, if such a PPT algorithm does not exist, then you do have a PRF.
So, ultimately, the question is asking whether or not you can construct a PPT algorithm that can distinguish the outputs of $F'$ from outputs of a random function. If you can construct such an algorithm (fairly easy in this case), then you have proved that it is not a PRF.
Let's suppose, however, that the problem were setup differently and in fact $F'$ were a PRF. How would you prove that? You can't go through every possible PPT algorithm and show that it cannot distinguish outputs of $F'$ from outputs of a random function. Instead, you would assume that $F'$ is not a PRF and show how this would lead to a conclusion that $F$ is also not a PRF. In other words, a proof by contradiction.