We show at first that Miku can play in such a way that Manni can never remove a rook. Let there be one rook in each rank and file in the initial configuration. Suppose that Manni moves a rook from square $(x, y)$ to square $(x, z)$. As a consequence, each file contains one rook but there are no rooks in rank $y$ and two rooks in rank $z$; let a rook be on square $(t, z)$, $t \neq x$. Let Miku move the rook from square $(t, z)$ to square $(t, y)$. After that, there is one rook in each rank and file again. In the case of Manni's move in the perpendicular direction, Miku's reply would be analogous. In such a way, Miku can reply to all Manni's following moves.

Now we show that also Manni can play so that Miku never wins. If there are two or more rooks on one rank or file in the initial configuration then Manni can remove one of them and win immediately. Therefore assume that initially there is one rook on each rank and file. Let there be rooks on squares $(1, u)$ and $(2, v)$. Manni can move the rook from $(2, v)$ to square $(2, u)$. As Miku is not allowed to remove this rook but Manni threatens to remove the rook on $(1, u)$ on the next move, Miku must move this rook to some square $(1, w)$. If $w \neq v$, Manni can win by moving the rook from square $(2, u)$ to square $(2, w)$ and removing either of the other rooks on rank $w$ on the next move. If $w = v$, then there is one rook in each rank and file again and Manni can continue with the same strategy.