Let's analyze the given problem step by step.

The function provided is $f(x, y) = \ln(y - x^2)$.

1. Domain

The domain of a logarithmic function is defined where its argument is positive, i.e., $\ln(z)$ is only defined for $z > 0$.

So, for $\ln(y - x^2)$, we need: $y - x^2 > 0 \quad \Rightarrow \quad y > x^2$ Thus, the domain is all points $(x, y)$ where $y > x^2$, as stated in the image.

Domain: $D = \{(x, y) \mid y > x^2 \}$

2. Range

The range of $f(x, y)$ is determined by the possible output values of the logarithmic function. Since $\ln(z)$ can take any real value when $z > 0$, the range of $f(x, y)$ is: $R = (-\infty, \infty)$ This matches the information in the image.

3. Level Curves

The level curves of the function $f(x, y) = \ln(y - x^2)$ are found by setting $f(x, y) = c$ for a constant $c$, which gives: $\ln(y - x^2) = c \quad \Rightarrow \quad y - x^2 = e^c \quad \Rightarrow \quad y = x^2 + e^c$ This equation represents a family of parabolas, where the constant $e^c$ shifts the parabola vertically.

Thus, the level curves are parabolas, as indicated.

Would you like more details or further explanations on any of these points?

Here are 5 related questions to expand on this:

1. What would happen to the domain if the function were $\ln(x^2 - y)$ instead?
2. How would the range change if we replaced the logarithm with a square root function?
3. Can you find a specific level curve for a given value of $c$, for instance, $c = 0$?
4. How do the level curves change if the function becomes $f(x, y) = \ln(1 + y - x^2)$?
5. How does the behavior of this function change near the boundary of the domain?

Tip: Always ensure the argument of a logarithmic function remains positive to avoid undefined values.