Inverse of a differentiable and bijective function $f: R \to R$ is also differentiable. Inverse of a differentiable and bijective function $f: R \to R$ is also differentiable.
True or false.
I know the statement is false.
If we take the function $x^{3}$ then its inverse is not differentiable at $0$.
If we think in terms of graph, the graph of a differentiable and bijective function should be smooth for all reals.
If the graph of  $f$ is smooth in $R$, then graph of $f^{-1}$ is also smooth (also bijective), since we just need to rotate the graph of $f$.
At first, I solved this problem using this graphical approach, and the statement seemed to be true.
What is wrong with graphical approach$?$
If the graph of a function is smooth in $R$, is it necessarily differentiable$?$

Your answer to the question "Is it true or false that the inverse of a bijective differentiable function is always differentiable?" is correct. It is false, and $y=x^3$ is a counter-example, because the inverse is not differentiable at $x=0$.
It is tempting to equate differentiable with "smooth" and indeed the term smooth is often used for a function which is infinitely differentiable. But if we are thinking graphically, a "smooth" curve does not necessarily represent a differentiable function, because the corresponding function might have "infinite derivative" at a point (and we normally regard that as meaning it is not differentiable at that point). 
If a function $f$ is bijective, then we can get the graph of its inverse by reflecting the graph in the $45^o$ line, so that the $x$-axis and $y$-axis are interchanged. Clearly, as a curve the graph is unchanged, but a zero gradient can become an infinite gradient, so that the inverse function is not necessarily differentiable (although there will usually just be a few isolated points where it has infinite derivative).
