Let $U \subset \mathbb{R}^2$ be open and $f\in C^1(U;\mathbb{R}^2)$. We also have $f(x_0)=y_0$. Suppose that $\overline{B}^2_\delta (x_0) \subset U$, so $\partial \overline{B}^2_\delta (x_0) = (\delta \cos(t),\delta \sin(t))+x_0$, when $t\in [0,2\pi]$. Now we set $\varphi(t)-x_0=(\delta \cos(t),\delta \sin(t))$.
How to derive the following
$$(f(\varphi(t))-y_0))\times (f(\varphi(t))-y_0)'=[\det f'(x_0)+o(|\delta|)](\varphi(t)-x_0)\times (\varphi(t)-x_0)', \space \space as \space \space \delta \rightarrow 0,$$
where $a\times b=(0,0,a_1b_2-a_2b_1).$
I do understand that $(\varphi(t)-x_0)\times (\varphi(t)-x_0)'=(0,0,1)$, but I don't understand why $(f(\varphi(t))-y_0))\times (f(\varphi(t))-y_0)' \rightarrow (0,0,\det f'(x_0))$, as $\delta \rightarrow 0$. I guess there is something simple that I don't see.
You have (Taylor on the first part, derivative on the second and inserting $0=(x_0)'$): $$ (f(\phi(t))-y_0)\times (f(\phi(t))-y_0)'= ((f'(x_0)+o(\delta))(\phi(t)-x_0)) \times (f'(x_0)(\phi(t)-x_0)')= [\det(f'(x_0))+o(\delta)] (\phi(t)-x_0)\times (\phi(t)-x_0)'$$ (using $\det AB=\det A \det B$ on two by two matrices [the last component of the cross product)