Recall. Define higher-order differentials. Let $p \geq 1,$ let $V$ be open in $\mathbf{R}^{n},$ let $\boldsymbol{a} \in V,$ and let $f: V \rightarrow \mathbf{R} .$ We shall say that $f$ has a pth-order total differential at $a$ if and only if the $(p-1)$ st-order partial derivatives of $f$ exist on $V$ and are differentiable at $a$, in which case we shall use the notation
$$D^{(p)} f(\boldsymbol{\alpha}, \boldsymbol{h}):=\sum_{i_{1}=1}^{n} \cdots \sum_{i_{p}=1}^{n} \frac{\partial^{p} f}{\partial x_{i_{1}} \ldots \partial x_{i_{p}}}(\boldsymbol{a}) h_{i_{1}} \cdots h_{i_{p}}, \quad \boldsymbol{h}=\left(h_{1}, \ldots, h_{n}\right) \in \mathbf{R}^{n}$$
Therefore,
$$ D^{(2)} f((a, b) ;(h, k))=h^{2} \frac{\partial^{2} f}{\partial x^{2}}(a, b)+2 h k \frac{\partial^{2} f}{\partial x \partial y}(a, b)+k^{2} \frac{\partial^{2} f}{\partial y^{2}}(a, b) . $$
Question- Find the second total differential of $f(x, y)=(x y)^{2}$.
My attempt: $$ D^{(2)} f((x, y) ;(h, k))=2 y^{2} h^{2} (a,b) +8 x y h k (a,b) +2 x^{2} k^{2} (a,b) $$
What I will do $(a,b)$'s in the my attempt?
$$ D^{(2)} f((a, b) ;(h, k))=h^{2} \left(\frac{\partial^{2} f}{\partial x^{2}}\right)_{(x=a, y=b)}+2 h k \left(\frac{\partial^{2} f}{\partial x \partial y}\right)_{(x=a, y=b)}+k^{2} \left(\frac{\partial^{2} f}{\partial y^{2}}\right)_{(x=a, y=b)} . $$ $D^{(2)} f((a, b) ;(h, k))$ means that the total differential is wanted a the point $(x=a,y=b)$, not at any point $(x,y)$. If the total differential was wanted at $(x,y)$ one would have written $D^{(2)} f((x, y) ;(h, k))$ . $$f(x,y)=x^2y^2 \quad\begin{cases} \frac{\partial^{2} f}{\partial x^{2}}=2y^2 \\ \frac{\partial^{2} f}{\partial x \partial y}=4xy \\ \frac{\partial^{2} f}{\partial y^{2}}=2x^2 \end{cases} \quad\begin{cases} \left(\frac{\partial^{2} f}{\partial x^{2}}\right)_{(x=a,y=b)}=2b^2 \\ \left(\frac{\partial^{2} f}{\partial x \partial y}\right)_{(x=a,y=b)}=4ab \\ \left(\frac{\partial^{2} f}{\partial y^{2}}\right)_{(x=a,y=b)}=2a^2 \end{cases} $$
$$ D^{(2)} f((a, b) ;(h, k))=h^2(2b^2) +2h k (4ab) + k^{2} (2a^2) $$
$$ D^{(2)} f((a, b) ;(h, k))=2b^2h^2 +8abh k + 2a^2k^2 $$ There is no $x$ and no $y$ into it because the total differential is evaluated at a specified point $(x=a,y=b)$.
The general formula would be : $$ D^{(2)} f((x, y) ;(h, k))=2y^2h^2 +8xyh k + 2x^2k^2 $$