Note that $a=0$ is special. Apart from not being defined at $0$, our function is then the same as the function $g(x)=x$, and has no local max or min.
We deal now with $a\gt 0$ and $a\lt 0$. I would deal with these separately.
Case $a\gt 0$: The derivative is $0$ at $x=0$ and at $x=2a$. We examine each of these points in turn.
(i) The bottom of the derivative is safely positive. Consider the top $x(x-2a)$ very near $x=2a$. Note that $x$ is positive. For $x$ near $2a$ but smaller than $2a$, we see that $x-2a$ is negative. For $x\gt 2a$, it is positive. so the derivative is negative for $x$ smaller than $2a$, but near $2a$, and positive afterwards. Thus our function near $2a$ is going down and then up, we have a local minimum at $x=2a$.
(ii) We now look at the derivative near $x=0$. If $x\lt 0$, then $x$ is negative. Also, $x-2a$ is negative there, so the derivative is positive. For $x\gt 0$ but close to $0$, we have $x$ positive and $x-2a$ negative, so the derivative is negative. Thus near $x=0$ our function is going up and then down, we have a local max at $x=0$. But the problem only asks about the interval $(0,\infty)$, so we needn't have bothered.
Case $a\lt 0$: We are only asked for local max and min in $(0,\infty)$. In the case $a\lt 0$ there are no critical points in $(0,\infty)$, so no local max or min. without that restriction, the analysis would go along lines similar to the case $a\gt 0$.
Best Answer
It sounds like the exercise is asking you to explain that there are no local minima of $f$, providing you with the graph of $f'$. If this interpretation of the exercise is correct, consider that candidates for local extrema of $f$ are at zeros of $f'$. There is one zero of $f'$ in the middle of the picture, but no sign change in $f'$ occurs there, so this is not a local extremum. (Near this point the graph would look something like the graph of $y=-x^3$ near $x=0$.)