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?
 A: 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 $f(x)=x$.
A: For $x\neq0$ and $x\neq1$ we obtain: $$f(x)=x^2f\left(\frac{1}{x}\right)=x^2f\left(\frac{1}{x}-1+1\right)=x^2\left(f\left(\frac{1}{x}-1\right)+1\right)=$$
$$=x^2+x^2f\left(\frac{1-x}{x}\right)=x^2+x^2\cdot\frac{f\left(\frac{x}{1-x}\right)}{\frac{x^2}{(x-1)^2}}=x^2+(x-1)^2f\left(\frac{x}{1-x}\right)=$$
$$=x^2+(x-1)^2f\left(\frac{1}{1-x}-1\right)=x^2+(x-1)^2\left(f\left(\frac{1}{1-x}\right)-1\right)=$$
$$=x^2-(x-1)^2+(x-1)^2\cdot\frac{f(1-x)}{(1-x)^2}=2x-1-f(x-1).$$
Thus, $$f(x)+f(x-1)=2x-1$$ or
$$f(x)+f(x+1)=2x+1,$$ which gives
$$f(x)+f(x)+1=2x+1$$ or
$$f(x)=x.$$
Now, show that $f(0)=0$ and $f(1)=1.$
