Consider a system of ellipses $E_u$ with same foci $(\pm u,0)$, here $u>0$ is a constant. In other words, we consider a system of equations $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a^2-b^2=u^2.$$
Q: How to show that any ellipse in $E_u$ satisfies $$xyy'^2+(x^2-y^2-u^2)y'-xy=0,$$ here $y'=\frac{dy}{dx}$?
We have $$\frac{2x}{a^2}+\frac{2yy'}{b^2}=0$$ or $$y'=-\frac{b^2x}{a^2y}.$$ Thus, we need to prove that $$\frac{xy\cdot b^4x^2}{a^4y^2}-\frac{(x^2-y^2-a^2+b^2)b^2x}{a^2y}-xy=0$$ or $$x(b^2-a^2)(b^2x^2-a^2y^2-a^2b^2)=0$$ or $$(b^2-a^2)x\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}-1\right)=0.$$ Done!