Proving that $y=x$ is an asymptote. I'm currently investigating the curve implicitly defined by $x^2-y^2= \ln (xy)$. Now I can see that by considering the region $xy>1$ and $xy<1$, we can determine where the curve is above or below the line $y=x$.
However, I do not feel that this is not enough to prove that the line $y=x$ is an asymptote in itself, just whether it approaches it from above or below.
For an implicitly defined curve like this, what kind of approach would I take to show that $y=x$ is an asymptote?

You could separate the variables:
$$x^2 - \ln x = y^2 + \ln y.$$
It's easy then to show that $y\to\infty$ as $x\to\infty$.
Take the ratio of the two sides:
$$\frac{y^2 + \ln y}{x^2 - \ln x} = 1.$$
For large positive $x$, you have large positive $y$ and $y^2/x^2 \approx 1$, 
so $y/x \approx 1$.
You can even write
\begin{align}
\frac{y^2}{x^2}
& = \left( \frac{y^2}{y^2 + \ln y} \right) \left( \frac{x^2 - \ln x}{x^2}\right)
 \frac{y^2 + \ln y}{x^2 - \ln x} \\
& = \left( \frac{y^2}{y^2 + \ln y} \right) \left( \frac{x^2 - \ln x}{x^2}\right)
\end{align}
Each of the quantities in parentheses goes to $1$ as $x\to\infty$.
