Points on $(x^2 + y^2)^2 = 2x^2 - 2y^2$ with slope of $1$ Let the curve in the plane defined by the equation: $(x^2 + y^2)^2 = 2x^2 - 2y^2$
How can i graph the curve in the plane and determine the points of the curve where $\frac{dy}{dx} = 1$.
My work:
First i found the roots of this equation with a change of variable $z = y^2$ and get: 

and then i tried to graph the point $ x - y$ and $x + y $ but i stuck i can't graph this and find the point  where the derivative is 1. Some help please.


I would try to solve this as an implicit derivative:
$$\frac{d}{dx}(x^2 + y^2)^2 = \frac{d}{dx}(2x^2 - 2y^2)$$
$$2(x^2+y^2)(2x+2y*y') =4x-4y*y'$$
$$2 x^3+2x^2 y* y'+2 y^3 y'+2 x y^2 = 4x-4y*y'$$
$$2x^2y*y' + 2y^3y' + 4y*y' = 4x-2xy^2-2x^2$$
$$y'(2x^2y + 2y^3+4y) =4x-2xy^2-2x^2$$
$$y' = \frac{4x-2xy^2-2x^2}{2x^2y + 2y^3+4y} = 1$$
$$\implies 4x-2xy^2-2x^2 = 2x^2y + 2y^3+4y$$
