A function is defined by this formula:
$$f:\mathbb N \times \mathcal P(\mathbb N) \to \mathbb N \times \mathcal P(\mathbb N) \\ f(x,y) = \{\min(\{x\} \cup y), y \setminus \{x\} \}$$ Find $$f^{-1}[\{0\} \times \mathcal P(\mathbb N)]$$
As for me, this question is not clear - if I understand correctly, I need to find all possible pairs $(x,y)$ such that $f(x,y) = (0, n)$ where $n \subseteq \mathbb N$. Therefore, I can think of two possibilities:
a) Zero comes from the first coordinate
b) Zero comes from the second coordinate or from both of them at the same time
a) Answer: $$\{(0, n) \mid n\subseteq \mathbb N\}$$
b) Answer:
$$\{(m,n) \mid m\in\mathbb N \land n\subseteq \mathbb N \land 0 \in n \}$$
Now, it's enough to take the sum of these two sets and we are done.
Is this the correct solution to this problem?
You have the right idea, but write it down properly, the result has to be a subset of $\mathbb{N} \times \mathcal{P}(\mathbb{N})$:
\begin{aligned} f^{-1}(\{0\} \times \mathcal{P}(\mathbb{N})) &= \{ (x,y) \in \mathbb{N} \times \mathcal{P}(\mathbb{N}) : (\min(\{x\}\cup y),y\setminus\{x\}) \in \{0\}\times\mathcal{P}(\mathbb{N})\}\\ &=\{(x,y) \in \mathbb{N} \times \mathcal{P}(\mathbb{N}) : \min(\{x\}\cup y) = 0\}\\ &=\left(\{0\} \times \mathcal{P}(\mathbb{N})\right) \cup \left(\mathbb{N} \times \mathcal{A}\right) \end{aligned}
where $\mathcal{A}:=\{ B \in \mathcal{P}(\mathbb{N}) : \{0\} \in B)$. Note that $y\setminus\{x\} \in \mathcal{P}(\mathbb{N})$ in any case, that gives you the second equality.