Suppose that $f:=(u,v):\Bbb R\to \Bbb R^2$ is $C^2$ and $(x_0,y_0)=f(t_0)$
A) prove that if $f'(t_0)\not=0$ then $u'(t_0)$ and $v'(t_0)$cannot both be zero.
B) if $f'(t_0)\not=0$ show that either there is $C^1$ function $g$ such that $g(x_0)=t_0$ and $u(g(x))=x$ for $x$ near $x_0$ or there is $C^1$ function $h$ such that $h(y_0)=t_0$ and $u(g(y))=y$ for $y$ near $y_0$
I guess, I need to use Implicit Function thm. But I dont know how to use this therem to solve these two parts. I just have been starting to learn this theorem by myself. Please explicitly show me. Thank you :)
This is the inverse function theorem.
For part A, you have to show that $f'(t)$ is just the vector consisting of $(u'(t), v'(t))$. You can show this quickly from the definition of the derivative, but it depends how your book defines it. For instance, if the definition for $v$ is the derivative of $f(t)$ is
$$\lim_ {t\to t_o} \left \| \frac{f(t) - f(t_0)} {t} -v \right \|=0,$$
then it just requires knowing that a vector goes to zero iff it goes to zero in each component.
Part B is really just the definition of the implicit function theorem. A function $f:\mathbb{R} \to \mathbb{R}^2$ can be thought of as two functions $f_1(t)$ and $f_2(t)$, which are the functions in the components. Read what the criteria are for applying the inverse function theorem, and apply it!