What Are the Level Curves for f(x,y)=e^-(2x^2+2y^2)?

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I am being asked to calculate level curves for the following equation:
f(x,y)=e^-(2x^2+2y^2) but I do not know where to start. Any advice on first steps would be greatly appreciated.
 
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Set e^-(2x^2+2y^2) = c, where c is a constant. Solve for x or y.

Once you do that, see if you can come up with a more general solution.
 
One obvious point: if e^{-(2x^2+ 2y^2)}= c then -(2x^2+ 2y^2)= ln(c) so that x^2+ y^2= -ln(c)/2 which is only possible if -ln(c)< 0 which means 0< c< 1. What figures will level curves be?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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