= Solution
Let $p=(x_0,y_0)$, $a=f_x(p)$ and $b=f_y(p)$. For sufficiently small $h,k$, decompose the increment into an $x$-segment at height $y_0+k$ and a $y$-segment at $x_0$. The one-variable <mean value theorem> applies on each segment because existence of its partial derivative gives continuity along the segment. Thus, when both increments are nonzero,
$$
f(x_0+h,y_0+k)-f(p)=h f_x(x_0+\theta h,y_0+k)+k f_y(x_0,y_0+\eta k)
$$
for some $0<\theta,\eta<1$. The same formula with the absent segment omitted covers zero increments. Continuity of both <partial derivatives> at $p$ makes the remainder after subtracting $ah+bk$ at most $\varepsilon(|h|+|k|)$ for small increments, and hence at most $\sqrt2\varepsilon\sqrt{h^2+k^2}$. Therefore \b[the <Fréchet derivative> is $(h,k)\mapsto ah+bk$]. Continuity throughout the disc is not required; continuity at the point suffices.
For the <determinant>, expand by <multilinearity of the determinant> in the columns of $I+H$. Taking no perturbed column gives one; taking exactly one gives $\operatorname{tr}H$; all other terms have degree at least two in the entries of $H$. Since matrix-entry functionals are bounded in any finite-dimensional norm, their sum is $O(\|H\|^2)$. Hence
$$
\det(I+H)=1+\operatorname{tr}H+O(\|H\|^2),\qquad\boxed{D\det_I(H)=\operatorname{tr}H.}
$$
For invertible $A$, <multiplicativity of the determinant> gives $\det(A+H)=\det A\det(I+A^{-1}H)$. The bounded linear map $H\mapsto A^{-1}H$ transfers the preceding remainder estimate, giving the <derivative of the determinant>
$$
\boxed{D\det_A(H)=\det A\operatorname{tr}(A^{-1}H).}
$$
To justify an open invertible neighborhood, let $\|K\|<1$. Completeness and submultiplicativity make the <Neumann series> $S=\sum_{j=0}^\infty(-K)^j$ converge in matrix norm. Multiplying finite partial sums by $I+K$ gives $I-(-K)^{N+1}$, which tends to $I$. Continuity of multiplication proves $(I+K)S=S(I+K)=I$. Thus \b[$I+K$ is invertible], and the series gives $(I+H)^{-1}=I-H+O(\|H\|^2)$ near zero.
Apply the first-derivative formula at $I+H$ to a test matrix $K$:
$$
\begin{aligned}
D\det_{I+H}(K)&=[1+\operatorname{tr}H+O(\|H\|^2)]\operatorname{tr}[(I-H+O(\|H\|^2))K]\\
&=\operatorname{tr}K+\operatorname{tr}H\operatorname{tr}K-\operatorname{tr}(HK)+O(\|H\|^2\|K\|).
\end{aligned}
$$
The last estimate is uniform for $\|K\|\le1$, so it proves differentiability of the derivative in the operator norm, rather than merely a directional second derivative. The <second derivative of the determinant> is the symmetric <bilinear map>
$$
\boxed{D^2\det_I(H,K)=\operatorname{tr}H\operatorname{tr}K-\operatorname{tr}(HK).}
$$
Its symmetry follows from $\operatorname{tr}(HK)=\operatorname{tr}(KH)$.
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