<s>[INST] Prove that $f(x)=x$ if the following holds true Let $f\colon\mathbb R \to \mathbb R$ be a continuous odd function such that 
1) $f(1+x)=1+f(x)$
2) $x^2f(1/x)=f(x)$ for $x\ne0$. 
Prove that $f(x)=x$.
 [/INST] If we do it in a more "functional equation"-ish approach, we notice that
$$\left(\frac{x+1}{x}\right)^2f\left(\frac{x}{x+1}\right) = f\left(\frac{x+1}{x}\right)$$
On the other hand,
$$\begin{align}f\left(\frac{x}{x+1}\right) &= f\left(1 - \frac{1}{x + 1}\right)\\
&= 1 + \left(-\frac{1}{x+1}\right)\\
&= 1 - \left(\frac{1}{x+1}\right)\\
&= 1 - \frac{f(x+1)}{(x+1)^2}
\end{align}$$
Substituting back:
$$\left(\frac{x+1}{x}\right)^2\left(1 - \frac{f(x+1)}{(x+1)^2}\right) = f\left(\frac{x+1}{x}\right)$$
$$\begin{align}\frac{1}{x^2}\left((x+1)^2 - f(x+1)\right) &= f\left(1 + \frac{1}{x}\right)\\
&= 1 + f\left(\frac{1}{x}\right) \\
&= 1 + \frac{f(x)}{x^2}\end{align}$$
Dividing throughout by $\frac{1}{x^2}$ (valid since $x \neq 0$):
$$\begin{align}(x+1)^2 - f(x+1) &= x^2 + f(x)\\
x^2 + 2x + 1 - x^2 &= f(x) + f(x + 1)\\
2x + 1 &= 1 + 2f(x)\end{align}$$
$$2x = 2f(x)$$
Which proves the desired statement:
$$f(x) = x$$
This is a rather long winded approach and I'm pretty sure it can be simplified a bit.
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