working with the set $\{(1,2),(2,1),(3,1),(5,4)\}$
I always second guess myself and sometimes confuse myself over simple things. The question asks whats the pre-image of $f^{-1}(1)$.
My answer is that the pre-image is $(2,3)$ since $2,3$ in domain both map to $1$ in the codomain
The pre-image of a function $f^{-1}(y) = \{x:f(x)=y\}$. In this case since $(2,1)$ and $(3,1)$ are in the function then the pre-image of $1$ is $\{2,3\}$.