Is there any fast method (for manual computation by hand) to find the determinant of the matrix
$$[D \alpha ]_{i,j} = \begin{cases} 1 + (\frac{\partial f}{\partial x_i})^2 & ; i = j \\ \frac{\partial f}{\partial x_i} \cdot \frac{\partial f}{\partial x_j} & ; i \not = j \end{cases}, \text{where} \quad i = 1,...,n$$ and $f: \mathbb{R}^n \to \mathbb{R}^n$
The matrix is equal to $$ I + \nabla f \cdot \nabla f^T, $$ where $\nabla f$ is a column vector, whose determinant equals $$ \det(I + \nabla f \cdot\nabla f^T) = \det( 1+ \nabla f^T \nabla f) = 1+ \nabla f^T \nabla f= 1 + \|\nabla f\|_2 ^2. $$ See Sylvester's determinant identity