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?

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 $f(1)=1.$
