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.


For determining the graph is better plot it as Lemur says in polar coordinates and see what happen at different angles with $r$. 
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For the points with derivative equal to one:
$$(x^2 + y^2)^2 = 2x^2 - 2y^2$$
$$x^4+y^4+2x^2y^2=2x^2-2y^2$$
$$4x^3+4y^3\frac{dy}{dx}+4xy^2+4x^2y\frac{dy}{dx}=4x-4y\frac{dy}{dx}$$
Setting $\frac{dy}{dx}=1$
$$4x^3+4y^3+4xy^2+4x^2y=4x-4y$$
