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 $f(x)=x$.
 A: 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.
A: Hint: (i) As $x$ is odd, what can you say about $f(0) = -f(-0)$?
(ii) Using 1), compute $f$ on the integers (use induction).
(iii) Using 2) compute $f$ for numbers of the form $\frac 1n$, $n \in \mathbb N - \{0\}$.
(iv) Using 1) again, this gives $f$ for rational numbers.
(v) Now use continuity to conclude.
A: This is some kind of addendum to a Hint in the answers. 

The function is odd and therefore it is zero at zero: $f(0)=0$. Therefore from two equalities in the question, we respectively get $f(n)=n$ and $f(\frac 1n)=\frac 1n$ for all integers $n$. 
For even numbers, we have $f(\frac n2)=\frac n2$ and $f(\frac 2n)=\frac 2n$. For odd numbers we have the following:
$$
f(\frac n2)=f(\frac{n-1}2+\frac{1}2)=\frac{n-1}2+\frac{1}2=\frac n2
$$
Therefore for all $n$, we have $f(\frac n2)=\frac n2$ and therefore $f(\frac 2n)=\frac 2n$. So far we can say that for all $0 \leq r\leq 2$, we have:
$$
f(\frac nr)=\frac nr \\
f(\frac rn)=\frac rn
$$
Now we can use induction to prove the previous equalities for all positive integers $r$. Suppose that for all $0\leq r\leq m-1$ we have:
$$
f(\frac nr)=\frac nr \\
f(\frac rn)=\frac rn
$$
We want to prove that :
$$
f(\frac nm)=\frac nm \\
f(\frac mn)=\frac mn.
$$
There are two numbers $q$ and $0 \leq r<m$ such that, we can write the integer $n$ as $qm+r$. We can then write the following:
$$
f(\frac nm)=f(\frac{qm}m+\frac{r}m)=q+f(\frac{r}m)=\frac nm
$$.
where the last equality comes from the assumption of induction, namely that for $0\leq r\leq m-1$ and all $n$ we have $f(\frac{r}m)=\frac rm$. This finishes the induction proof and says that for all integers $m$ and $n$: 
$$
f(\frac nm)=\frac nm \\
f(\frac mn)=\frac mn.
$$
The function is identity function over all rational $x$. For irrational number $x$, we consider a sequence of rational numbers $r_n$ converging to $x$ and then use the continuity argument to show $f(x)=x$.
