<s>[INST] 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.

 [/INST] 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 $\frac{dy}{dx}=1$
$$4x^3+4y^3+4xy^2+4x^2y=4x-4y$$
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