Procedure to find a level curves I'm having trouble finding level curves. What's the procedure? In this case, for example:
$z=x^2+y^2=k$ $\hookrightarrow y=\sqrt(k-x^2)$
Then I sketch this based on knowing the formulas of circunference, parabola, ellipse, etc? Or do I sketch this based on first, second derivatives and general procedures of graph sketching?
Or every formula will lead to common things like hyperbole, parabola, circunferences,etc?
What should I focus my study on? Right now I'm not good in solving level curves problems...
Like finding level curves for this: $$ z=e^{x^2-y^2}$$
Thanks in advance.
 A: In your first example, the proper solution is
$$y=\pm \sqrt{k-x^2}$$
You left out the plus-or-minus. That is not a small thing: there are usually two values of $y$ for each $x$, and that greatly affects the plotting of the curves.
I would say that there is no single general method for finding level curves, in a similar way that there is no general way to solve a fifth-degree polynomial and no general way to find the antiderivative of a function. There are some techniques that sometimes work, but nothing always works. The general problem is just too complicated.
In a given problem, you should first look to see if it does reduce to a familiar problem such as a parabola, ellipse, etc. If not, you can use the derivatives to sketch the curve. Don't forget to try some particular values of $x$ or $y$ to find a few key points on the curve.
In your last example, if $z$ is constant then so is $x^2-y^2$. This should look familiar, as the level curves for that are hyperbolas.
If you want to improve, first get comfortable with the conic sections (points, lines, double lines, circles, ellipses, parabolas, and hyperbolas). You could then move on to more complicated ones.
A: Either you use a computer, or you sketch the curves based on recognizing the equation as something you already know.
In the first case you should recognize $x^2+y^2=k$ as the equation for a circle with radius $\sqrt k$ rather than try to rewrite it to get $y$ as a function of $x$.
Similarly for $e^{x^2-y^2}=k$ you would first take the logarithm on both sides to get $x^2-y^2=\log k$ and then recognize the latter as a hyperbola with asymptotes $x=\pm y$ -- or, when $\log k=0$, the crossing lines $x=\pm y$ themselves.
Of course you're not always going to get something you recognizing. In that case it is either the computer, or a lot of labor with pencil and paper to find enough points to connect them freehand.
