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.


You've already got some good suggestions on the graphing part using polar coordinates (if you are familiar with that). For the other, implicitly differentiate the given relation to obtain ${dy\over dx}$:
\begin{align*}
(x^2 + y^2)^2 &= 2x^2 - 2y^2\\
2(x^2+y^2)(2x+2y{dy\over dx})&=4x-4y{dy\over dx}\\
2x^3+2x^2y{dy\over dx}+2xy^2+2y^3{dy\over dx}&=2x-2y{dy\over dx}\\
(2x^2y+2y^3+2y){dy\over dx}&=2x-2x^3-2xy^2\\
{dy\over dx}&={2x-2x^3-2xy^2\over 2x^2y+2y^3+2y}\\
{dy\over dx}&={x(1-x^2-y^2)\over y(x^2+y^2+1)}\\
\end{align*}
Now solve ${dy\over dx}=1$ in terms of $x$ and $y$, keeping in mind you can use the original expression to simplify things along the way. Edit: This was much more involved than I anticipated; see below.
Here's the situation (note ${dy\over dx}$ is not defined at the origin):

So how to find these coordinates? As noted in another answer, the original relation can be expressed in polar coordinates as $r^2=2\cos(2t)$. Recall that the slope of the tangent line in polar coordinates is given by 
$$
{dy\over dx}={{dr\over dt}\sin t+r\cos t\over {dr\over dt}\cos t-r\sin t},
$$
so setting ${dy\over dx}=1$ and using $r^2=2\cos(2t)\implies {dr\over dt}={-2\sin(2t)\over r}$, we get
\begin{align*}
{dy\over dx}&=1\\
{dr\over dt}\sin t+r\cos t&={dr\over dt}\cos t-r\sin t\\
{-2\sin(2t)\over r}\sin t+r\cos t&={-2\sin(2t)\over r}\cos t-r\sin t\\
-2\sin(2t)\sin t+r^2\cos t&=-2\sin(2t)\cos t-r^2\sin t\\
r^2&={2\sin(2t)[\sin t-\cos t]\over \cos t+\sin t}\\
2\cos(2t)&={2\sin(2t)[\sin t-\cos t]\over \cos t+\sin t}\\
\end{align*}
Multiply on the right by ${\cos t-\sin t\over \cos t-\sin t}$, use $\cos^2t-\sin^2t=\cos(2t)$ and $\sin(2t)=2\sin t\cos t$ and rearrange to get 
\begin{align*}
\sin(2t)&=-(\sin t+\cos t)^2\\
2\sin t\cos t&=-1-2\sin t\cos t\\
4\sin t\cos t&=-1\\
\sin(2t)&=-{1\over 2}\\
t&={7\pi\over 12},\ {11\pi\over 12}.
\end{align*}
But $r^2=2\cos(2t)$ so, $t={11\pi/12}\implies r^2=2\cos(11\pi/6)=\sqrt{3}$. Thus, the polar coordinates of this point are $(3^{1/4},11\pi/12)$ which in rectangular coordinates is $$(3^{1/4}\cos(11\pi/12),3^{1/4}\sin(11\pi/12))\approx (-1.27,0.34)$$ as shown in the figure above.
(The fourth quadrant point of tangency can be found similarly.)
