This uses plain TikZ with intersections library for the intersection point. Note that scope environment can be nested, several \clips are used for filling. If you want the rectangle with more rounded corners, then simply use [rounded corners=3mm] e.g., in the definition of \bb.
I believe that pgfplots also can be easily drawn this, with suitable options in the axis environment. However, plain TikZ makes me more comfortable.

\documentclass[tikz,border=3mm]{standalone}
\usepackage{amsmath}
\usetikzlibrary{intersections}
\begin{document}
\begin{tikzpicture}[scale=2]
\def\bb{[rounded corners] (-1,-1) rectangle (2,1.5)}
\def\curveA{plot[domain=-1:2,smooth,samples=100] (\x,{\x/(sqrt(1+\x*\x))})}
\def\curveB{plot[domain=-1:2,smooth,samples=100](\x,{pow(\x,4)-\x})}
\begin{scope} \clip \bb;
\begin{scope}
\clip \curveA|-cycle;
\clip \curveB--cycle;
\fill[cyan!20] \bb;
\end{scope}
\draw (-1,0)--(2,0) (0,-1)--(0,1.5);
\draw[magenta,thick,name path=A] \curveA;
\draw[cyan,thick,name path=B] \curveB;
\path[name intersections={of=A and B}] (intersection-2) node[cyan,rectangle,minimum size=2mm,draw,thick]{};
\end{scope}
\draw[cyan] \bb;
\foreach \i in {-.5,0,...,1.5} \draw (\i,.05)--(\i,-.05);
\foreach \j in {-.5,0,...,1} \draw (.05,\j)--(-.05,\j);
\path
(-1,0) node[left]{$-1$}
(2,0) node[right]{$2$}
(0,-1) node[below]{$-1$}
(0,1.5) node[above]{$1.5$}
(1.2,-.5) node{$y=x^4-x$}
(.6,1.2) node (N) {$y=\dfrac{x}{\sqrt{x^2+1}}$};
\draw[-stealth] (N)--+(-60:.6);
\end{tikzpicture}
\end{document}
axis line style = {rounded corners). I also proposed a new title of the question. Feel free to reverse my proposal. – Dr. Manuel Kuehner Feb 03 '20 at 21:54rounded corners. If there is not an answer I close my question :-) thank to your link. – Sebastiano Feb 03 '20 at 21:57