Hint: The substitution $y = tx$ makes it possible to find the parametrization of the entire curve with the parameter $t$.
Using the hint I got the following parametrization:
$$x(t) = t^2 - 1 \\ y(t) = t^3 - t $$
Here's the plot of the curve from Wolfram Alpha
I think the relevant formula here is $|D| = \int_{t_0}^{t_1}|y(t) x'(t)| \,dt$ but I'm not certain what are the correct values for $t_0$ and $t_1$.
For $t=0$ we have $(x(0), y(0)) = (-1, 0)$ and for $t=1$ we have $(x(1), y(1)) = (0, 0)$. Since the figure is symmetric about the $x$-axis we have:
$$2\int_0^1 (t^3 - t) 2t \,dt = - \frac{8}{15} $$
But area obviously cannot be negative.
Flipping the region of integration gets us $-2\int_1^0 (t^3 - t) 2t \,dt = \frac{8}{15} $ but why is integrating from $1$ to $0$ (instead of $0$ to $1$) the correct thing to do here? Is it correct in the first place? And, of course, is $\frac{8}{15}$ the correct result?
The parametrisation is rather unnecessary. It is clear that $$y^2=x^2(x+1)$$ has only two zeros, at $x=0$ and $x=-1$. Since the curve is symmetric about $x$-axis, the loop is formed in the region $-1\leq x\leq 0$. Hence, required area would be: $$A=2\int_{-1}^0 |y|\ dx=2\int_{-1}^0 \sqrt{x^3+x^2}\ dx=2\int_{-1}^0| x|\sqrt{x+1}\ dx=- 2\int_{-1}^0 x\sqrt{x+1}\ dx$$ Which can be easily evaluated using $u=x+1$. The result comes out to be $\frac {8}{15}$ only, as expected.
You have done pretty much the same thing, but you did not take modulus of the function, which you should have done since scalar area is asked here.