How can I clear $y$ in this funtion to sketch a level curve? I have $f(x,y) = \ln\left(\frac{1-\sqrt{x^2-y^2}}{2xy}\right)$ and the exercise tell me to find three level curves.
I know that $f(x,y)=K$ will give me the level curves so:
$\ln\left(\frac{1-\sqrt{x^2-y^2}}{2xy}\right)=K$
$\frac{1-\sqrt{x^2-y^2}}{2xy}=e^K$
$1-\sqrt{x^2-y^2}=2xye^K$
$1-2xye^K=\sqrt{x^2-y^2}$
$(1-2xye^K)^2=x^2-y^2$ 
And I got stuck, this looks like an hyperbola equation but I dont know how to clear the $y$ to make the graphs of the level curves.
 A: Are you sure you're supposed to graph them, not just provide their equations? And if you do have to graph them, are you sure you're required to do that by hand, not with some technology?
Your work so far is more or less correct, although I wouldn't simplify the equation as far as you did. The line $\frac{1-\sqrt{x^2-y^2}}{2xy}=e^K$ is okay, but going further is not quite correct. The problem is that you acquire a lot of extraneous points which were not in the domain of the original function. For example your next line obtained by multiplying by $2xy$ allows $x$ and $y$ to be zero, but in the original that wasn't in the domain. And then squaring is even worse…
Anyways, since the question says to give three level curves, all you've gotta do now is pick three different values for $K$ (any values you like), and you'll have three equations of three level curves. For example, if we choose $K=0$, we'll get the equation $\frac{1-\sqrt{x^2-y^2}}{2xy}=1$, which is one of those level curves. If your task is to provide equations of three level curves, something like this would be good enough.
And no, they are not hyperbolas. A hyperbola can have an equation $x^2-y^2=\operatorname{const}$ (for example), but in this equation the left-hand side is not constant at all. More importantly, hyperbolas are given by quadratic equations in $x$ and $y$, but this function is not quadratic — if you expand the left-hand side, you'll see a polynomial of degree four. So these are some degree four curves, whose shapes can be quite complicated — see for example how it looks for $K=0$ after squaring (so it's not correct because it's too much, but it gives a better idea about its shape) and before squaring. This is why I seriously doubt you were required to actually draw them.
