Prove inverse function theorem (1 dimensional case) 
Let $I$ be an open interval in $\mathbb{R}$ that contains point $a$
   and let $f:I\rightarrow \mathbb{R}$ be a continuously differentiable
   function such that $f'(a)\ne 0$. Then there exist open intervals
   $U\subset I$ and $V$ in $\mathbb{R}$ such that restriction of $f$ to
   $U$ is a bijection of $U$ onto $V$ whose inverse $f^{-1}: V\rightarrow
 U$ is differentiable.

Since $f'(a)$ is non zero $f'$ is continuous, it must be the case that $f'$ does not change sign in some small ball $U=(a-\epsilon, a+\epsilon)\subset I$. Thus $f$ is strictly increasing (decreasing) on that ball. Therefore $f$ restricted to $U_0$ is a bijection from $U_0$ onto $V=f(U)$. Thus there is $f^{-1}:V\rightarrow U$ which is a continuous bijection. We can use mean value theorem, so for $x,x_1 \in V$ such that $x\ne x_1$ there is some $\theta$ between $f^{-1}(x)$ and $f^{-1}(x_1)$ such that
$$x-x_1=f(f^{-1}(x))-f(f^{-1}(x_1))=(f^{-1}(x)-f^{-1}(x_1)) f'(\theta)$$
thus $$\frac{f^{-1}(x)-f^{-1}(x_1)}{x-x_1}=\frac{1}{f'(\theta)}$$
(we know that $f'(\theta)$ is not zero)
And here is where I'm stuck. Any help please?

Let $x$ and $x+h$ belong to the domain of $f$ where inverse exists and $f(x)=y$ and $f(x+h)=y+k$
Also $f^{-1}(y+k)=x+h$ and $f^{-1}(y)=x$
Limit as $h$ tends to zero $\frac {f^{-1}(y+k)-f^{-1}(y)}{k}$
Now $f^{-1}(x+h)$ is $y+k$ and $f^{-1}(x)$ is $y$
so the limit becomes
$= \frac {h}{f(x+h)-f(x)}$ and now take h to the denominator and apply limit
Sorry I am bad at typing do it is difficult to write an answer
