We have
$$f(x,y)=z=\ln(x-y)$$
For some level curve where $z=k$, we then have
$$k=\ln(x-y)$$
Here, there are only two variables, $x$ and $y$. It is now possible to write $x$ as a function of $y$, and vice versa. You should have a set of two-dimensional functions. All you have to do is graph them.
I won't give you the answer in that case, but I can give you a different example. Say we have a different function, and it turns out that
$$k=e^{x^2-y}$$
We can then write
$$y=x^2-\ln(k)$$
and graph this function for each $k$ that is given to us.
Now that you've drawn up $y=x$, try graphing level curves of $f(x,y)$. Which side of that line do they fall on?
This should show you that all of your $k$s fall within the domain of the function. In fact, you can prove that for any real $k$, the level curve will be on the same side of $y=x$.
So I did some more looking around at similar problems on the internet, as well as looked at some other books, and I've come up with an answer I feel is much less shady. I'll post it here if anyone has a comment on it, but also so that if anyone else has a similar question to mine perhaps this answer will be helpful.
So I will basically take the advice of one of the comments above and just solve the PDE.
We have our original function $f(x,y) = x^2y^3$ whose level curves we wish to be perpendicular to. As such we set up the function $g$ and take its level curve at $g(x,y)=0$, as so: $g(x,y) = L$. We need them to intersect orthogonally, i.e. their gradients are orthogonal. So
$$
\nabla f \perp \nabla g \\
\iff \Big(\nabla f \Big| \nabla g\Big) = 0 \\
\iff 2xy^3\frac{\partial g}{\partial x} + 3x^2y^2\frac{\partial g}{\partial y} = 0
$$
Note that if $x = 0$ or $y = 0$ the level curve of $f(x,y) = K$ is just a point, so we can safely assume $x,y\neq 0$. Hence the above becomes
$$
2y\frac{\partial g}{\partial x} + 3x\frac{\partial g}{\partial y} = 0
$$
This holds when $\frac{\partial g}{\partial x} = \frac{1}{2y}$ and $\frac{\partial g}{\partial y} = -\frac{1}{3x}$. We integrate both of these:
$$
g(x,y) = \frac{x}{2y} + \phi(y) \\
g(x,y) = -\frac{y}{3x} + \psi(x)
$$
Where $\phi$ and $\psi$ are some arbitrary functions of one variable. Combining these two equations for $g$ we find that $g(x,y) = \frac{x}{2y} - \frac{y}{3x} + C$. Since $g$ passes through $(2,-1)$ we have that $C= -5$. Finally we take the level curve when $g(x,y)=0$, which implies the curve is $3x^2-y^2=10$. //
I prefer this answer because it seems much less shady than the one in the book, but also it's quite nice since it actually reveals that there are an infinite amount of these level curves, which can be trivially found by setting $g(x,y) = L$ for different values of $L$. It should be noted that I'm very inexperienced with PDE's, so my solution might have some shady steps.
Best Answer
For $e^{-1}$ you get $$x^2 + (y-1)^2=1$$ which is a circle.
For $e^{-1/4}$ you get $$x^2 + (y-1)^2=1/4$$ which is another circle.
For $1=e^{0}$ you get $$x^2 + (y-1)^2=0$$ which is a point.