algebra precalculus - Functional equation - Cyclic Substitutions

Please help solve the below functional equation for a function $f: \mathbb R \rightarrow \mathbb R$:

$$\begin{align} &f(-x) = -f(x) , \text{ and } f(x+1) = f(x) + 1, \text{ and } f\left(\frac 1x\right) = \frac{f(x)}{x^2} \\ &\text{ for all } x \in \mathbb R \text{ and } x \ne 0 . \end{align}$$

I know this will be solved by cyclic substitutions, but I'm unable to figure out the exact working. Can someone explain step wise?

Answer

I don't know if this is what is meant with "cyclic substitutions", but it is a solution.

First, we observe that

$$f(0)=0, f(1)=1, f(-1)=-1$$
has to be true. Also, it suffices to determine $f(x)$ for $x>0$, because the rest follows from the condition $f(-x)=-f(x)$.

Let $x>0$.
Applying some of the conditions, we have

$$\begin{aligned} f(x)+1 &= f(x+1) \\ &= f(1(x+1)^{-1})(x+1)^2 \\ &= f(1-x(x+1)^{-1})(x+1)^2\\ &= (1+f(-x(x+1)^{-1}))(x+1)^2\\ &= (1-f(-x(x+1)^{-1}))(x+1)^2\\ &= (1-f(-x(x+1)^{-1}))(x+1)^2\\ &= (1-f((x+1)x^{-1})x^2(x+1)^{-2})(x+1)^2\\ &= (1-f(1+x^{-1})x^2(x+1)^{-2})(x+1)^2\\ &= (1-(1+f(x^{-1}))x^2(x+1)^{-2})(x+1)^2\\ &= (1-(1+f(x)x^{-2})x^2(x+1)^{-2})(x+1)^2\\ &= (x+1)^2-x^2-f(x)\\ \end{aligned}$$

This yields $f(x)=x$.