Question: Describe the level surfaces of the following function: $f(x, y, z) = x^2 − y^2 + z^2$
I'm currently trying find the surface in which the function represents but I can't seem to figure out. I'm leaning into thinking it's a elliptic cone with its format as there's two positives. Anyone have a clue?



This is almost the canonical equation of a hyperboloid, so this is what the surface will be. Some intuition on the three types of hyperboloid is provided below. The TL; DR is that you can rotate a hyperbola either about its major axis or about its conjugate axis (the axis between the two sheets), giving rise to two different kinds of hyperboloid.
The level curves verify $$x^2-y^2+z^2=A$$ for some $A$, or equivalently $$y^2-r^2=-A$$ where $r^2=x^2+z^2$. This means that $y$ depends on $x$ and $z$ only through $r$, which is the distance from the point $(x, y, z)$ to the $Y$ axis. Therefore, the surface is rotationally symmetric about the $Y$ axis.
There are three cases depending on the sign of $A$. Nice plots of the three possibilities have been provided by Raffaele in their answer.
If $A$ is positive, let $A=a^2$, and we have $$\frac{r^2}{a^2}-\frac{y^2}{a^2}=1$$ which is the equation of a hyperbola with asymptotes $r=\pm y$ and vertices at $(r=\pm a, y=0)$. The surface is this hyperbola rotated about the $Y$ axis.
If $A$ is $0$, then we have the degenerate case $y^2= r^2$, or $y=\pm r$, which is a cone along the $Y$ axis.
If $A$ is negative, let $A=-a^2$, and we have $$\frac{y^2}{a^2}-\frac{r^2}{a^2}=1$$ which is the equation of a hyperbola with asymptotes $r=\pm y$ and vertices at $(r=0, y=\pm a)$. This is the same hyperbola as in the positive $A$ case, but with the $Y$ and $R$ axes swapped. The surface is this hyperbola rotated about the $Y$ axis.