I am trying to draw the graph of the derivative of this function:
Since the function has a maximum at $x = 0$, the derivative at this point must be zero. In addition, the function is increasing up to $x = 0$ and decreasing from this value, so the derivative has $y> 0$ for $x <0$ and $y <0$ for $x> 0$. When $x$ tends to $+\infty$, the graph tends to zero, so $f'(x) = 0$. My problem is to understand the shape of the concavity, or if there is a concavity in the graph. The graph I made was as follows:
Could you tell me if the concavities and the shape of the graph are right and why do you have the concavities that way?
Your graph appears to look good. However, note that the derivative for values for $x \to -\infty$, the derivative does not reach an asymptote, as you assume, instead it goes to infinity. The slope is very big if $x \to -\infty$, as indicated by the graph.
So yes, it should grow "like $x^2$ up.", because the slope is very big. Additionally, the left side of the graph seems close to $y = ax^3 + 1$, whose derivative is $3ax^2$, a quadratic function.
The "trick" says things about the concavity of $f^\prime$. To find the concavity, you have to check the sign of second derivative of the function. In this case, the function is the first derivative of your original function. $(f^\prime)^{\prime \prime}$ = $f^{\prime \prime \prime}$.