<s>[INST] Non linear first order ODE in in $\mathbb{R}^2$ 2 Let $a, b \in \mathbb{R}$ and consider the system of ODEs 
$$\left\{ \begin{matrix} x' = ax+ by^2, & x(0) = x_0\\ y' = ay +bx^2, & y(0) = y_0  \end{matrix} \right. $$
where $x, y :\mathbb{R} \to \mathbb{C}.$
I wonder whether there is, in general, $F$ such that $F(x,y) =0.$ 
For example, if $a =0,$ and $b \neq  0,$ we can write 
$$\frac{dx}{dy} = \frac{y^2}{x^2},$$ which implies that 
$$x^3- y^3 + c = 0$$ for some constant $c.$  
Thank you for any hint.
 [/INST] $$\left\{ \begin{matrix} x' = ax+ by^2, & x(0) = x_0\\ y' = ay +bx^2, & y(0) = y_0  \end{matrix} \right.$$
Multiply first equation by $x^2$ and second equation by $y^2$
$$\left\{ \begin{matrix} x^2x' = ax^3+ by^2x^2 & . \\ y'y^2 = ay^3 +bx^2y^2 & .  \end{matrix} \right.$$
$$x'x^2-ax^3=y'y^2-ay^3$$
$$\frac 13 (x^3)'-ax^3=\frac 13 (y^3)'-ay^3$$
$$\frac 13( x^3-y^3)'=a(x^3-y^3)$$
You can integrate, substitute $z=x^3-y^3$
$$z'=3az$$
$$ \implies x^3(t)=y^3(t)+Ke^{3at}$$

You can try to integrate the differential equation with exactness 
$$\frac {dx}{dy}=\frac {ax+by^2}{ay+bx^2}$$
$$({ay+bx^2})dx-({ax+by^2})dy=0$$
$$Pdx+Qdy=0 \implies \partial_y P=a, \partial_x Q=-a $$ $$\implies \partial_y P -\partial_x Q=2a$$
Try an integrating factor $z=x^3-y^3$
And the formula for the integratinf factor $z=z(x,y)$ is 
$$\boxed {\frac {d\mu}{\mu}=\frac {\partial_y P- \partial_x Q}{Q \partial_x z -P \partial_y z}}$$
$$\frac {d\mu}{\mu}=\frac {2a dz}{Q \partial_x z -P \partial_y z}$$
$$\frac {d\mu}{\mu}=-\frac {2 dz}{3z}$$
$$\mu =\frac 1 {\sqrt[3]{z^2}}$$
$$\mu (x,y) =\frac 1 {\sqrt[3]{(x^3-y^3)^2}}$$
</s>