Consider the following ODE system:
$\frac{dx}{dt}=x^2+y^2-y\\ \frac{dy}{dt}=-2xy-x$
I want to prove that there exists a solution $(x(t),y(t))$ such that
$\lim_{t\to +\infty}(x(t),y(t))=\lim_{t\to -\infty}(x(t),y(t))=(0,0)$ and $(x(t),y(t))\neq (0,0)$ for any $t$.
I can prove that when $F(x,y)=6x^2y+3x^2+2y^3-3y^2$, we have $\frac{d}{dt}F(x(t),y(t))=0$. So if there exists a solution satisfying the condition, it should be on the curve of $6x^2y+3x^2+2y^3-3y^2=0$. Then I have no idea about what I can do next. Should I try to find a special solution?
Note that this answer finds the requested solution explicitly: (I am sure there are some clever arguments such no explicit solution is needed)
As the solution should approach $(x,y)=(0,0)$ with $F(0,0)=0$ at $t\to\pm \infty$, you are interested in finding all points with $F(x,y)=0$. The solution (of $x$ in terms of $y$) is given by (we take the branch with $x,y>0$ for convenience) $$ x = y \sqrt{\frac{\frac32-y}{\frac32 +3 y}} \tag{1}\,.$$ A potential solution of your ODE is on this curve. A nice initial value is $x=0, y= \frac32$ at $t=0$.
To find a solution, we reexpress the second ODE as $$ x = - \frac{y'}{1+2 y} \tag{2}\,.$$
Equating (1) and (2), we find $$ - \frac{y'}{1+2 y} = y \sqrt{\frac{\frac32-y}{\frac32 +3 y}}\,.$$ This, equation can be solved by separation of variables. We obtain $$ - \int_{3/2}^{y} \frac{dz}{z(1+2 z)}\sqrt{\frac{\frac32 +3 z}{\frac32-z}} = \int_{3/2}^{z} \frac{\sqrt{3} dz}{z \sqrt{3+ 4z - 4 z^2}} = t \,.$$ Evaluating the integral yields the implicit solution $$-\cosh^{-1}\left(\frac{\frac32 +y}{2y} \right) = t\,.$$
With this, we find the (explicit) solution $$ y= \frac{3}{2 \cosh(t) -1}, \qquad x= \frac32 \frac{ \sinh(t/2)}{\cosh(3 t/2)}$$ where $x$ was determined from (1).
You can check that $x,y \to 0$ for $t\to \pm \infty$.