Determine whether a contour $f(x,y) = C$ is convex First of all, three dimension is more complex so let's restrict to two dimension here. Suppose a given contour $f(x,y) = C$ is closed--I don't know how to check the closeness mathematically, but let's just assume it is. I did some experiment, three typical examples are shown below. 
By observation I have the following hypothesis about convexity if $f(x,y)$ is a polynomial:
all right hand side signs are positive, and all powers of $x$ and $y$ are even. This is supported by the counter examples in the middle and the right panel. But I am not sure if it is actually true.
A directly related question would be to generalize to arbitrary function $f(x,y)$.
Edit: I should add that what I mean is the convexity of the region within the contour. I guess the question is better changed to "determine whether the region inside a contour: $f(x,y)<=C$, is convex". I apologize for not expressing it clearly.
 A: 
I have the following hypothesis about convexity if $f(x,y)$ is a polynomial: all right hand side signs are positive, and all powers of $x$ and $y$ are even

Your examples all have $f(x, y) = 1$ with the top-degree coefficients positive, so let's restrict to this special case.
Assuming the sign condition refers to the coefficients of $f$ (i.e., on the left-hand side), your hypothesis might mean: 1. the sign condition implies convexity of the (sub-)level sets of $f$, 2. convexity of the sub-level sets of $f$ implies the sign condition, 3. both, or 4. something else.
In any event, neither condition implies the other: If
$$
f(x, y) = x^{2} + y^{2}(1 + cy(1 + y)),
$$
the sub-level set $\{(x, y) : f(x, y) \leq 1\}$ is approximately circular, hence convex for $|c|$ sufficiently small. This proves that a polynomial containing an odd-degree term (with mixed-sign coefficients or not) can have a strictly convex sub-level set.
Conversely, if
$$
f(x, y) = x^{2} + y^{2} + cx^{2}y^{2},
$$
the sub-level set $\{(x, y) : f(x, y) \leq 1\}$ is not convex for $|c|$ sufficiently large, proving that a polynomial containing only even-degree terms with positive coefficients need not strictly convex sub-level sets.
A: Let suppose $f$ is enough nice function, such that  we can locally rewrite the curve $f(x,y)=C$ as graph of $y=g(x)$ for some $g$ around the point $(x_0, g(x_0))$ such that   $f(x_0, g(x_0))=C$. We can do this taking into account implicit function theorem! Then it is easy to check that the graph of $y=g(x)$ is convex subset of $R^2$ if and only if $g$ is linear function, i.e. $y=ax+b$  !
A: If the questioner by convexity of the level curves $f(x,y)=C$ He/She means the convexity of the sublevel set $f(x,y) \le C$, this is actually an equivalent definition of Quasiconvex functions. There are some characterization of this kind of functions in literatures, but not exactly what OP wants (a charterization in terms of sign of coefficients and exponents). Take look at my example below, it shows why thinking about this kind of characterization is not helpful.
For any  real numbers $\alpha,$ $\beta$, $\lambda$ and $n > 0$ the function $ f(x,y)=(\alpha x + \beta y + \lambda)^{n} $ has above property, but the parameters $\alpha,$ $\beta$, $\lambda$ and $n$ are free to choose  !
An another category of Quasiconvex functions is $ f(x,y)=(\alpha x  + \lambda)^{n} + (\beta y + t)^m.$
