Player 0 wins with optimal play. In fact, we prove that Player 1 cannot prevent Player 0 from creating a row of all zeroes, a column of all zeroes, or a $2 \times 2$ submatrix of all zeroes. Each of these forces the determinant of the matrix to be zero. For $i,j=1, 2,3$, let $A_{ij}$ denote the position in row $i$ and column $j$. Without loss of generality, we may assume that Player 1's first move is at $A_{11}$. Player 0 then plays at $A_{22}$: $$\begin{pmatrix} 1 & * & * \\ * & 0 & * \\ * & * & * \end{pmatrix}$$ After Player 1's second move, at least one of $A_{23}$ and $A_{32}$ remains vacant. Without loss of generality, assume $A_{23}$ remains vacant; Player 0 then plays there. After Player 1's third move, Player 0 wins by playing at $A_{21}$ if that position is unoccupied. So assume instead that Player 1 has played there. Thus of Player 1's three moves so far, two are at $A_{11}$ and $A_{21}$. Hence for $i$ equal to one of 1 or 3, and for $j$ equal to one of 2 or 3, the following are both true: (a) The $2 \times 2$ submatrix formed by rows 2 and $i$ and by columns 2 and 3 contains two zeroes and two empty positions. (b) Column $j$ contains one zero and two empty positions. Player 0 next plays at $A_{ij}$. To prevent a zero column, Player 1 must play in column $j$, upon which Player 0 completes the $2 \times 2$ submatrix in (a) for the win.