If $f:R-\left\{\frac{{\displaystyle 3}}{{\displaystyle 7}}\right\}\to R-\left\{\frac{{\displaystyle 3}}{{\displaystyle 7}}\right\}$ is given by $f\left(x\right)=\frac{3x+5}{7x-3}$, then the statement which is not true

$\left(fof\right)\left(x\right)=f\left[f\left(x\right)\right]=\frac{3f\left(x\right)+5}{7f\left(x\right)-3}=\frac{3\left(\frac{3x+5}{7x-3}\right)+5}{7\left(\frac{3x+5}{7x-3}\right)-3}=\frac{3(3x+5)+5(7x-3)}{7(3x+5)-3(7x-3)}$

$=\frac{9x+15+35x-15}{21x+35-21x+9}=\frac{44x}{44}=x\Rightarrow$ option (2) is true

$\left(fof\right)\left(x\right)=x\Rightarrow f^{-1}\left(x\right)=f\left(x\right)\Rightarrow$ option (1) is true

$\left(fofofof\right)\left(x\right)=fof\left[\left(fof\right)x\right]=\left(fof\right)\left(x\right)=x\Rightarrow$option (4) is true