$f: \mathbb{R}^{2 \times 2} \cong\mathbb{R}^4 \longrightarrow \mathbb{R}$
$ f: A \mapsto \det A$
So let $(a_{11},a_{21},a_{12},a_{22})^T \in \mathbb{R}^4$
I need to
a) reason why $f\in C^2$.
$\det(A)=a_{11}a_{22}-a_{21}a_{12}$, which is a polynomial therefore $f\in C^\infty.$
b) determine $df(A) (\in L(\mathbb{R}^4,\mathbb{R}))$ ($L$ means linear mapping. $df$ is the gradient)
$df(A)=(a_{22},-a_{12},-a_{21},a_{11})$
c) determine $d^2f(A)(v_1,v_2)$ for $v_1,v_2 \in \mathbb{R}^4$
Are a) and b) correct? How is c) done?
a) and b) are correct, except that I would write the total derivative $df(A)$ as a row matrix $$df(A) = \begin{pmatrix} a_{22} & -a_{12} & -a_{21} & a_{11}\end{pmatrix}$$ because it represents a linear map $\mathbb R ^4 \to \mathbb R$ and acts on column vectors of $\mathbb R^4$ by matrix-vector multiplication.
Then for c), the Hessian of $f$ at $A$ is $$H = \begin{pmatrix} & & & 1 \\ & & -1 & \\ & -1 & & \\ 1 & & & \end{pmatrix},$$ whence the associated bilinear form is given by $$d^2 f(v_1,v_2) = v_2^t Hv_1 = v_1^4 v_2^1 - v_1^3 v_2^2 - v_1^2 v_2^3 + v_1^1 v_2^4,$$
where I wrote $v_i^j$ for the $j$th component of the vector $v_i$.