Derivative of inverse functions Show graphically that
$$
(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}. \\
$$
PS: I know that $f(x)$ and $f^{-1}(x)$ are symmetrical about the line $y=x$. It should be proved  by showing that multiplying the slopes of the tangents of points $x$ and $f(x)$ equals 1.
As you say, the graph of a function and its inverse are symmetrical about the line $y=x$ - or, one is a reflection about that line of the other.  This geometric fact corresponds to the algebraic fact that $y=f(x)$ if and only if $x=f^{-1}(y)$.  Again stated geometrically, the point $(x,y)$ is on the graph of $f$ if and only if the point $(y,x)$ is on the graph of $f^{-1}$.  To see why this interchange of $x$ and $y$ corresponds to reflection across the line $y=x$, refer to the following diagram:



If we move horizontally from the point $(x,y)$ to the line $y=x$, we don't change the $y$-coordinate and we arrive at a point whose $x$ coordinate is the same as its $y$ coordinate; thus we arrive at the point $(y,y)$.  Moving vertically, we don't change the $x$-coordinate; so moving vertically brings us to the point $(x,x)$ on the line $y=x$.  Reversing the process brings us to the point $(y,x)$. Thus, the point $(y,x)$ is on the opposite side of the line $y=x$ and along the diagonal of a square - exactly as the geometric reflection of $(x,y)$ would be.

We now ask - how does this interchange of $x$ and $y$ affect the slope of a line.  Given a line with points $(x_1,y_1)$ and $(x_2,y_2)$, it's slope is 
$$\frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}.$$
The corresponding line generated by interchanging $x$ and $y$ has slope 
$$\frac{x_2-x_1}{y_2-y_1},$$
which is exactly the reciprocal.  We can again illustrate this to make it clear:



Finally, we apply these observations to the graph of an invertible function $f$ together with it's inverse function $f^{-1}$ and the tangent lines at $(x,y)$ and $(y,x)$.
What does mean "graphically"? It is not clear to me. 

We know that $f(f^{-1}(x))=x$ by definition

Thus $[f(f^{-1}(x))]'=f'(f^{-1}(x))(f^{-1})'(x)=1$

And then $(f^{-1})'(x)=\dfrac{1}{f'(f^{-1}(x))}$

But this is not clearly "graphical"...