Let's analyze the mathematical problem from the image you uploaded.

The function provided is:

$f(x, y) = \ln(y^2 - 2x^2)$

1. Domain:

For the natural logarithm function $\ln(z)$, the argument $z$ must be positive. Therefore, we need to solve the inequality: $y^2 - 2x^2 > 0$

This represents a region of the plane where the function is defined. Specifically, this inequality describes a hyperbolic region.

So, the domain is:

$D = \{(x, y) \mid y^2 > 2x^2 \}$ which means the domain is the set of points $(x, y)$ where $y^2 > 2x^2$.

2. Range:

The natural logarithm function $\ln(z)$ can take any real value as long as its argument $z$ is positive. Since the argument $y^2 - 2x^2$ can become arbitrarily large (as $y \to \infty$ or $x \to 0$) and close to 0 (but still positive), the range of the function is:

$R = (-\infty, \infty)$

3. Level Curves:

The level curves of $f(x, y) = c$ correspond to the curves where:

$\ln(y^2 - 2x^2) = c$

This simplifies to:

$y^2 - 2x^2 = e^c$

This is the equation of a hyperbola for each value of $c$. Therefore, the level curves are hyperbolas.

Let me know if you'd like more details or have any specific questions!

Here are some related questions to deepen understanding:

1. How can we graphically represent the domain $y^2 > 2x^2$?
2. Why does the logarithmic function's argument need to be positive?
3. What properties of hyperbolas can we infer from the level curves of the function?
4. How does changing the value of $c$ affect the shape and position of the hyperbolas?
5. Could the function's domain include the boundary $y^2 = 2x^2$? Why or why not?

Tip: Always consider the domain of a function before solving or graphing it, especially when dealing with logarithmic or radical functions.