<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] I won't do the graph, just the algebra.
Implicit differentiation gives
$$2(x^2+y^2)(2x+2yy')=4x-4yy'$$
so setting $y'=1$ and simplifying leads to
$$(x^2+y^2)(x+y)=x-y\qquad(*)$$
If we now rewrite the equation for the curve as
$$(x^2+y^2)^2=2(x-y)(x+y)$$
we see that multiplying both sides of equation $(*)$ by $2(x+y)$ gives
$$2(x^2+y^2)(x+y)^2=2(x-y)(x+y)=(x^2+y^2)^2$$
Ignoring the point on the curve at $(0,0)$ (where the derivative is not defined), we can cancel a $(x^2+y^2)$ from the above, leaving $2(x+y)^2=x^2+y^2$, which simplifies to $x^2+y^2=-4xy$ (which will mean that one of $x$ and $y$ will have to be positive and the other one negative).  Plugging this into the original equation for the curve gives $(-4xy)^2=2x^2-2y^2$, or
$$y^2={x^2\over1+8x^2}$$
Plugging this into the original equation for the curve gives
$$\left(x^2+{x^2\over1+8x^2}\right)^2=2x^2-2{x^2\over1+8x^2}$$
Simplifying this (and keeping in mind we're already ignoring $x=0$) produces
$$16x^4-24x^2-3=0$$
The real roots for this are
$$x=\pm\sqrt{3+2\sqrt3\over4}\approx\pm1.27$$
The corresponding values for $y$ (recalling that $-4xy$ must be positive) are
$$y=\mp\sqrt{-3+2\sqrt3\over4}\approx\mp.34$$
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