As you have already found by using the partial derivatives, the double point of the curve $\mathcal{C}: y^2 = x^3 + 2x^2$ is at the origin $(0, 0)$. Now let me try to explain you a way to make use of this double point to find all the rational points in the curve. Basically the process will give you a rational parametrization of the coordinates of any point on the curve.
The idea is to "project" from the double point onto some rational line. The picture below shows the curve $\mathcal{C}$ in red, in which first of all, you can see the double point at the origin where the curve actually intersects itself.

Also, I've plotted the rational line $ x = 1$, which is the vertical line shown in blue. And to explain the idea of the projection I've plotted the general line shown in green in the picture.
The basic idea of the projection from the double point is the following. Since a line in the plane has an equation of the form $y = mx + b$ and the polynomial defining your curve is the cubic polynomial $f(x, y) := y^2 - x^3 - 2x^2$, then when you look at the intersection points of the line $y = mx + b$ and the curve, you will have to substitute this $y$ in $f(x, y)$ to get the polynomial $f(x, mx + b) = (mx + b)^2 - x^3 - 2x^2$, which is a cubic in $x$. So in general it has three roots, which will correspond to the $x$ coordinates of the intersection points of the line with the curve $\mathcal{C}$.
But now comes into play the importance of the double point at the origin. When you have a double point, that means that this point is counted twice in the intersection, that means geometrically that there is only one different point in the intersection of the line and the curve, as you can see in the picture.
Then the idea of the projection is to match up this point on the curve, with the corresponding point in the line, which in this case I choose to be the vertical line shown in blue, but you could pick any line you like with rational coefficients. Then from this you can find a rational parametrization of the curve, which in turn gives you all the rational points on the curve.
To be more explicit, any point on the blue line has the form $(1, t)$ for $t \in \mathbb{R}$. Then the equation of the line passing through the double point at the origin, and the point $(1, t)$ is $y = tx$. To find the intersection point different from the origin, we solve the system
$\begin{cases} y^2 = x^3 + 2x^2\\ y = tx \end{cases} \implies t^2 x^2 = x^3 + 2x^2 \implies x^3 + 2x^2 - t^2 x^2 = 0 $
$ \implies x^2 (x + 2 - t^2 ) = 0 \implies x = 0 \ \ \text{or} \ \ x = t^2 - 2 $
The solution $x = 0$, which has multiplicity 2, corresponds to the double point at the origin. The other solution $x = t^2 - 2$ corresponds to the other intersection point on the curve. Then since $y = tx$, this implies that $y = t(t^2 - 2)$.
And this is basically it, we have the correspondence between the points on the curve, and the points on the line $x = 1$ given by
$ \begin{array} \{ \text{points on the curve $y^2 = x^3 + 2x^2$} \} &\longleftrightarrow \{ \text{points on the line $x = 1$} \}\\ \quad \quad \quad \quad \quad \quad (t^2 - 2, t(t^2 - 2))& \longleftrightarrow \quad (1, t) \end{array} $
So all the rational points on the curve have the form $(t^2 - 2, t(t^2 - 2))$ for $t \in \mathbb{Q}$.
You should read at least the first chapter of Silverman and Tate's book Rational Points on Elliptic Curves, where they explain this idea beautifully.
I hope that this helps you a little.