$f(x,y)=x^2y^2$, $a=(2,-1)$
So I'm using this definition:
$$ \lim\limits_{h \to 0}\frac{\Vert f(a+h)-f(a)-Df(a)h\Vert}{\Vert h\Vert}=0 $$
I get $$ \lim\limits_{h \to 0}\frac{\Vert (2+h_1)^2(h_2-1)^2-4(1)-4h_1+8h_2\Vert}{\Vert h\Vert}\\ =\lim\limits_{h \to 0} \frac{\vert 4h_2^2+4h_2^2h_1-8h_1h_2-2h_1^2h_2+h_1^2h_2^2+h_1^2\vert}{\Vert h\Vert} $$
I know I want to use squeeze theorem somehow but the quotient I have looks pretty useless right now.
Use these inequalities: \begin{align*} &h_1^2 \le h_1^2 + h_2^2 = \|h\|^2\\ &h_2^2\le h_1^2+ h_2^2 \le \| h\|^2\\ & |h_1 h_2| \le \frac 12 (h_1^2 + h_2^2) = \frac 12 \|h\|^2 \end{align*} The sum you have in numerator is therefore bounded above by $$4\|h\|^2 + 2|h_2| \|h\|^2+ 4 \|h\|^2+|h_1| \|h\|^2+\frac 14 \|h\|^2 + \|h\|^2,$$ dividing by $\|h\|$, the whole thing goes to zero as $\|h\|\to 0$.