At (0, 0), the curve y^2=x^3+x^2

$y^2=x^3+x^2 \Rightarrow 2 y \frac{d y}{d x}=3 x^2+2 x$

$\Rightarrow \frac{d y}{d x}=\frac{3 x^2+2 x}{2 y}=\frac{3 x^2+2 x}{2 \sqrt{x^3+x^2}}=\frac{3 x+2}{2 \sqrt{1+x}}$

∴  $\left.\frac{d y}{d x}\right|_{(0,0)}=\frac{2}{2}=1 \Rightarrow \theta=45°$

∴  the curve bisects the angle between the axes.