I am trying to compute the Taylor expansion to second order of the following function about point $(x,y)=(a,a)$, with $a<1$, $$f(x,y)=(1-x)(1-y).$$
If I am not mistaken, this expansion can be given by
$$(1-a) (1-a)-(x-a)(1-a)- (y-a)(1-a),$$
where it can be seen that e.g. there are not terms in $(x-a)^2$, which I think it results from the fact that $\frac{\partial^2 f}{\partial x^2}=0$.
Is my analysis correct ? and if so, can I claim that the Taylor expansion to order $n \ge 2$ will be the same as for $n=1$?
Recall that the Taylor expansion of order $k$ around a point $p \in \mathbf R^d$ of a $C^k$-function $f \colon U \subseteq \mathbf R^d \to \mathbf R^{d'}$ is given by $$ T^p_kf(x)= \sum_{|\alpha|\le k} \frac{1}{\alpha!}\partial^\alpha f(p)(x-p)^\alpha $$ for $k = 2$ we have $$ T^{(a,a)}_2 f(x) = f(a,a) + \partial_x f(a,a)(x-a) + \partial_y f(a,a)(y-a) + \frac 12 \partial_x^2 f(a,a)(x-a)^2 + \partial_{x}\partial_{y}f(a,a)(x-a)(y-a) + \frac 12 \partial_y^2f(a,a)(y-a)^2 $$ We have \begin{align*} \partial_x f(x,y) &= -(1-y)\\ \partial_y f(x,y) &= -(1-x)\\ \partial_x\partial_y f(x,y) = 1\\ \partial_x^2 f = \partial_y^2 f &= 0 \end{align*} This gives $$ T^{(a,a)}_2f(x,y) = (1-a)(1-a) - (1-a)(x-a) - (1-a)(y-a) + (x-a)(y-a) $$ You are missing the $x$-$y$-term, note that $\partial_x\partial_y f\ne 0$. You are correct in saying, that there is no $(x-a)^2$ term due to $\partial_x^2 f = 0$. As $\partial^\alpha f = 0$ for indices $\alpha$ of order $\ge 3$, we have that all expansions of order $\ge 3$ are equal to the one for $k = 2$.