The functional equation $\frac{f(x)}{f(1-x)} = \frac{1-x}{x}$ One set of solutions to this is $f(x) = \frac{c}{x}$ for constant $c$. Are these the only solutions?
 A: Let $f$ be a solution to the equation, and let $g(x):=(x+ \frac{1}{2})f(x + \frac{1}{2})$. Then $g$ is an even function (i.e. 
$ g(x)=g(-x)$). Conversly, let $h$ be any even function. Then by setting  $f(x):=\frac{h\left(x-\frac{1}{2}\right)}{x}$, we get $xf(x)=(1-x)f(1-x)$. Thus the set of all solutions to this equation is exactly the set of functions of the form:
$$f(x):=\frac{h\left(x-\frac{1}{2}\right)}{x}$$
Where $h$ is an even function.

Some examples:
\begin{align*} 
f(x) &= \frac{c}{x} \\
\\
f(x) &= \frac{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}{x} = 1-x \\
\\
f(x) &=\frac{\cos\left(x-\frac{1}{2}\right)}{x} \\
\end{align*}
A: No, since $f(x)=1-x$ is also a solution, as I wrote in the comments.
A: Among degree one polynomials, you can find solutions. Suppose $f(x)=a+bx$. Then you require
$$\frac{a+bx}{a+b(1-x)}=\frac{1-x}{x}$$
$$(a+bx)x=(1-x)(a+b-bx)$$
$$ax+bx^2=a+b -bx-(a+b)x+bx^2$$
$$0 = (a+b)(1-2x)$$
so that $b=-a$. You therefore have a family of functions
$$f_a(x)=a\cdot(1-x)$$
which are solutions.
You can try the same procedure with other types of functions with arbitrary coefficients and see what you might get.
