$x^y+y^x=c$ across the reals Inspired by: For integers $a\ge b\ge 2$, is $f(a,b) = a^b + b^a$ injective? (a question worth looking at IMHO).

My question is very simple. Let $x,y,c$ be positive real numbers, where $c$ is constant. $x^y+y^x=c$ represents a curve in 2D. Can anyone figure out a way of writing $y=f(x)$, not necessarily with standard functions? How about $x=f(t)$, $y=g(t)$? Is it possible? Any attempt I made rapidly disintegrated into a horrible mess...
To make clearer the second part of the question: is it possible to figure out if there is a way or not?
 A: In brief, no, I do not believe it is possible. You have enumerated three of the most common ways of writing formulas for curves in the plane: 


*

*implicitly, e.g. $x^y + y^x = c$; 

*explicitly, e.g. $y=f(x)$; 

*parametrically, e.g. $x=f(t)$, $y=g(t)$. 


Sometimes only one is reasonable/possible/worth one's effort. For this one, I'm pretty sure that an implicit formula is the only one that is possible.

One might attempt to address your clarified second part, but you'll have to say what kind of "way" you will accept, what kind of "formula" $y=f(x)$ is satisfactory to you. Unless you specify the allowable formulas, your question is too vague to have an answer. Let me explain by giving a few alternate scenarios.
Scenario #1: You are happy with an abstract existence theorem. In that case, by applying the implicit function theorem for any particular point $p=(a,b)$ on the curve, if $\frac{\partial}{\partial y}(x^y + y^x)$ is nonzero at $p$, then there does exist of a function $y=f(x)$ which parameterizes a small portion of the curve near $p$. Similarly, if $\frac{\partial}{\partial x}(x^y + y^x)$ is nonzero at $p$ then there exists a local parameterization $x=f(y)$ at $p$. This is a powerful abstract method for proving the existence of parameterizaitons of curves. However, it does not yield an explicit formula for $y=f(x)$, it is merely an existence theorem.
Scenario #2: You want a very concrete polynomial function $y=a_n x^n + \cdots + a_1 x + a_0$. In that case it can be proved, again with some calculus, that no such function exists. I am pretty sure that no rational function will work either, and that no algebraic function will work either, but the proofs will get harder and harder. As you attempt to prove stronger and stronger negative results like this it is quite likely at some point you will hit an unsolved problem.
Scenario #3: A function is proved to exist by some abstract theorem, no-one has a name for it, and someone then names it. And if it is sufficiently useful, it becomes standard. Take the function $f(x) = \int_1^x \frac{1}{t} dt$, for example, whose existence is guaranteed by the fundamental theorem of calculus. At some point someone (apparently Mercator) realized that this function was the inverse function of $y=e^x$, and so it was a kind of logarithm and hence was very useful, and eventually someone dreamed up the name "natural logarithm" for this function, and along the way it became one of the "standard functions" (apologies for the "fake history").
