I want to show that 1 is a regular value of $\det:GL(n,\mathbb{R})\rightarrow\mathbb{R}$, i.e. that for all $A\in GL(n,\mathbb{R})$ the rank of $D(f\circ\phi^{-1})(\phi(A)):\mathbb{R}^{n^{2}}\rightarrow\mathbb{R}$ is maximal (and hence equal to 1), where $\phi$ is a chart on $GL(n,\mathbb{R})$ at $A$.
I'm guessing I should prove that $D(f\circ\phi^{-1})(\phi(A))$ is surjective, but I don't know how to do this because I don't have an expression for $D(f\circ\phi^{-1})(\phi(A))$. I do know that $\mbox{det}_{*}(I)(X)=\text{tr}(x)$ but I don't know how to go from here.
We can use the formula we have for computing the determinant by expanding it along any row or column, namely if we expand along the $i$-th row we obtain:
$det(X) = \sum_{j=1}^n(-1)^{i+j}a_{ij}m_{ij}$
Where $m_{ij}$ is the determinant of the $i,j$-th minor of $X$. We can see that the partial with respect to the $i,j$-th entry will then be $\frac{\partial det}{\partial a_{ij}} = (-1)^{i+j}m_{ij}$. Now all that is left to check is that if the matrix has a nonzero determinant, the jacobian will have rank at least 1. Can you take it from here?
A cool fact, you can use this along with the regular value theorem to prove that $SL_n(\mathbf{R})$ is a Lie subgroup $GL_n(\mathbf{R})$!