Find a solution for $f\left(\frac{1}{x}\right)+f(x+1)=x$ Title says all. If $f$ is an analytic function on the real line, and $f\left(\dfrac{1}{x}\right)+f(x+1)=x$, what, if any, is a possible solution for $f(x)$?
Additionally, what are any solutions for $f\left(\dfrac{1}{x}\right)-f(x+1)=x$?

Just looking at $f(1/x)+f(x+1)=x$. My assumption is that you mean a function defined and analytic at every point on the real line. If so, then there is no such function. 
You would have
$$
\begin{align}
\lim_{x\to\infty}f(x)&=\lim_{x\to0^+}f(1/x)\\
&=\lim_{x\to0^+}\left(x-f(x+1)\right)\\
&=-f(1)
\end{align}
$$
And then:
$$
\begin{align}
f(0)&=\lim_{x\to0^+}f(x)\\
&=\lim_{x\to\infty}f(1/x)\\
&=\lim_{x\to\infty}\left(x-f(x+1)\right)&\Big(\lim_{x\to\infty}f(x+1)&=\lim_{x\to\infty}f(x)=-f(1)\Big)\\
&=\lim_{x\to\infty}(x)+f(1)
\end{align}
$$
So $f(0)$ does not exist. Now, if you wanted a function that is analytic on the positive reals, then that would be something else.
