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?

The chain rule can be used to show the derivative of the inverse is the inverse of the derivative. By composition of inverse functions we have $f(f^{-1}(x) ) = x . $ Differentiating both sides of the equation and applying the chain rule to the left hand side yields $f'(f^{-1}(x))\cdot \frac{d}{dx}(f^{-1}(x))=1.$ Then dividing gives $\frac{d}{dx}(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}.$
