The folium of Descartes: Given the function $F(x, y)= x^3+y^3-3xy$, the set $L={ \{(x,y) \in \mathbb{R}^2: F(x,y)=0} \}$ and $q_0=(x_0,y_0)=(\frac{3}{2}, \frac{3}{2})\in M$, how can I find an ordinary differential equation together with the initial condition.
My idea: first implicitly differentiate the given equation.
Result: $\frac{dy}{dx} = \frac{2y-x^2}{y^2-2x}$. Is that correct? What is the next step?
As Jyrki noted in a comment, you have the right idea, but the correct DE is $$y'(x)=\frac{x^2-y}{x-y^2}. $$ As for an initial condition, just use $y(3/2)=3/2:$ the point on the folium. At this point, I'd say you were done.