No stationary points means $$3x^2+2px+p\neq 0$$
Use the discriminant of this quadratic equation $D=(2p)^2-4\cdot 3\cdot p$. In order for a quadratic equation to have no real roots the discriminant has to be negative. Hence, $4p^2-12p<0$ or $4p(p-3)<0$.
Can you complete it from here?
EDIT:
A product of two real numbers can only be negative if only one of them is negative. As $p>0$ from the beginning you can conclude $p-3<0$ or $p<3$. Combined with $p>0$ we get the result $0<p<3$.
Your approach will work for some ODEs, but not all. For instance, the ODE
$$xy''+y'+y=0$$
has one singular point at $x=0$, which your method would find. However, the ODE
$$xy''+xy+\tan(x)y=0$$
has a singularities at $x=0$ and $x=\pi/2+\pi n$. Your method would not find the latter. Why?
The formal definition for singularities starts by rewriting your ODE in the form
$$y''+\frac{Q(x)}{P(x)}y'+\frac{R(x)}{P(x)}y=0$$
We then ask if the functions $\frac{Q(x)}{P(x)}$ and $\frac{R(x)}{P(x)}$ are analytic. If you aren't familiar with the term analytic, it essentially means that the functions have an infinite number of well-defined derivatives (you can take derivatives forever). If either of those functions are not analytic at some point $x_0$, we call that point a singular point. In the first example, we have
$$\frac{Q(x)}{P(x)}=\frac{R(x)}{P(x)}=\frac{1}{x}$$
which is not analytic at $x=0$ since it blows up to $\infty$. In the second example, we have
$$\frac{Q(x)}{P(x)}=1$$
and
$$\frac{R(x)}{P(x)}=\frac{\tan(x)}{x}$$
Note that $\tan(x)$ blows up to $\pm\infty$ at $x=\pi/2+\pi n$. Thus, there are singularities at $x=\pi/2+\pi n$. I suggest you read up on analytic functions if this isn't clear to you. Once you've found all the singular points, all other points are ordinary points.
Best Answer
The calculations are correct but your sketch is wrong because (0,0) isn't a turning point there.