Rationality of the Lemniscate. This question is exercise 2 of Chapter 4 in Kunz' textbook of algebraic curves.

Let $f$ be the lemniscate with equation
  $$(X^2 + Y^2 )^2 = α(X^2 − Y^2) \;\; (\alpha \in K^\times )$$
and let $x, y \in K[f]$ be the associated coordinate functions. Prove the
  rationality of $f$ by showing that $R(f) = K(t)$ with $t := \frac{x^2+y^2}{x-y}.$

Kunz defines a function as rational if $\mathcal{R}(f)\cong K(t)$ where $\mathcal{R}(f)$ is the ring of rational functions on $f$, and $t$ is transcendental over $K$.
So far I have not got far. I have tried parametrizing in $t$, however I ended up with trig functions. I have also tried the map $K[t]\to \mathcal{R}(f)$ which maps $t\mapsto\frac{x^2+y^2}{x-y}$ but I couldn't show that this extends to a map on $k(t)$. Thanks for any advice.

In the case $K=\mathbb{R}$ (where we can graph the lemniscate) the way to parametrize the lemniscate is by using the circles $x^2+y^2=t(x-y)$. You can check that:
$$
\begin{split} &&x^2+y^2=t(x-y)\\ &\implies& \left(x-\frac{t}{2}\right)^2+\left(y+\frac{t}{2}\right)^2=\frac{t^2}{2} \end{split}
$$
so that the parametrization is by circles centered at $(t/2,-t/2)$ with radius $t/\sqrt{2}$. Graphing some of these circles, you can see that they intersect the lemniscate at the origin (with multiplicity 2) and also at one other point. It is this other point that will give us the parametrization. We first check where the lemniscate intersects these circles using the substitution $x^2+y^2=t(x-y)$ in the equation for the lemniscate:
$$
\begin{split} && t^2(x-y)^2=a^2(x^2-y^2) \\ &\implies& t^2(x-y)^2=a^2(x-y)(x+y) \\ &\implies& t^2(x-y)=a^2(x+y) \\ &\implies& (t^2-a^2)x=(t^2+a^2)y \\ &\implies& y=\frac{t^2-a^2}{t^2+a^2}x\end{split}
$$
Where we have divided by $x-y$ since the only solution when $x=y$ is the $(0,0)$ point which we already know about (and is not the point we are interested in). This last equation defines a line so all we need to do is check where this line intersects the original circle (other than the origin, $x=0$). Given $ y=\frac{t^2-a^2}{t^2+a^2}x$, the equation for the circle becomes:
$$
\begin{split} && x^2+y^2=t(x-y) \\ &\implies& x^2+\left(\frac{t^2-a^2}{t^2+a^2}x\right)^2  =t\left(x-\frac{t^2-a^2}{t^2+a^2}x\right) \\ &\implies& \frac{(t^2+a^2)^2+(t^2-a^2)^2}{(t^2+a^2)^2}x^2=t\left(\frac{t^2x+a^2x-t^2x+a^2x}{t^2+a^2}\right) \\ &\implies& \frac{(t^2+a^2)^2+(t^2-a^2)^2}{t^2+a^2}x =2a^2t \\ &\implies& x=\frac{2a^2t(t^2+a^2)}{t^4+2a^2t^2+a^4+t^4-2a^2t^2+a^4} \\ &\implies& x=\frac{a^2t(t^2+a^2)}{t^4+a^4} \end{split}
$$
Substituting this in to get $y$ yields:
$$
y=\frac{t^2-a^2}{t^2+a^2}x=\frac{a^2t(t^2-a^2)}{t^4+a^4}
$$
This gives the required parametrization of the lemniscate in terms of $t$.
