Find all continuous functions $f(x+1)=f(x)+1$ and $f\left(\frac1x\right)=\frac1{x^2}f(x)$ 
Find all continuous functions $f:\mathbb R \rightarrow \mathbb R$ such that
1) $\forall x \in \mathbb R$
  $$f(-x)=-f(x)$$
2) $\forall x \in \mathbb R$
  $$f(x+1)=f(x)+1$$
3$\forall x \in \mathbb R/ \{0\}$
  $$f\left(\frac1x\right)=\frac1{x^2}f(x)$$

My work so far:
1) $f(0)=0$
2) $\forall x \in \mathbb Z$ $$f(x+m)=f(x)+m$$
3) I see that the answer $f(x)=x$, but can not rove.

Get $y = f(x)$ , then i have 
$$f(x+1)=y+1,f(\frac{1}{x+1})=\frac{y+1}{(x+1)^2},f(\frac{-1}{x+1})=\frac{-(y+1)}{(x+1)^2}$$
$$f(\frac{x}{x+1})=f(-\frac{1}{x+1}+1)=\frac{-(y+1)}{(x+1)^2}+1=\frac{x^{2}+2x-y}{(x+1)^2}$$
$$f(\frac{x}{x+1})=f(\frac{1}{\frac{x+1}{x}})=\frac{f(\frac{x+1}{x})}{(\frac{x+1}{x})^2}=\frac{\frac{x^{2}+y}{x^2}}{(\frac{x+1}{x})^2}=\frac{x^2+y}{(x+1)^2}$$
So i have 
$$\frac{x^{2}+2x-y}{(x+1)^2}=\frac{x^2+y}{(x+1)^2}$$
$$x=y$$ 
And easy to check $f(0)=0,f(-1)=-1$ so 
$$f(x) \equiv x$$
