curve of $(x^2+y^2)^2=2(x^2-y^2)$ The diagram shows the curve $(x^2+y^2)^2=2(x^2-y^2)$ and one of its maximum points $M$. Find the coordinates of $M$. 
My attempt.
Differentiate the equation and I got 
$\frac{dy}{dx}=-\frac{x(x^2+y^2-1)}{y(x^2+y^2+1)}$
How should I proceed?

HINT.-You have a closed curve, symmetrical with respect to the two axes (a lemniscate indeed) and you have the alternative of take the explicit form for $y$ solving a quadratic equation; you have
$$y^4+2(x^2+1)y^2+(x^4-2x^2)=0$$ By symmetry you can take only the positive signs in solving so you have 
$$y=\left(-x^2-1+\sqrt{4x^2+1}\right)^{\frac 12}$$
$$y'=\frac{-x+\frac{2x}{\sqrt{4x^2+1}}}{\sqrt y}$$
Hence you get for the maximun $$x=\frac{\sqrt 3}{2}$$ Thus $$y=\frac 12$$
