what value of a is f(f(x))=x ?

it will be related to x

as we get - a² -ax -1=0 or x=0

$\textbf{Here $\mathbf{f(x)=\frac{ax}{x+1}}$}$\\\\ $\textbf{Now $\mathbf{f(f(x))=x\Leftrightarrow f(x)=f^{-1}(x).................................(1)}$}$\\\\ $\textbf{Now We have to Calculate $\mathbf{f^{-1}(x)}$ Using $\mathbf{f(x)}$}$\\\\ $\textbf{So $\mathbf{f^{-1}(x)=\frac{x}{a-x}}$ (Calculate yourself.)}$\\\\ $\textbf{Now Put value of $\mathbf{f^{-1}(x)}$ in equa.......(1),we Get $}$\\\\ $\mathbf{\frac{ax}{x+1}=\frac{x}{a-x}\Leftrightarrow x^2.(1+a)+x.(1-a^2)=0=0.x^2+0.x}$\\\\ \textbf{So We Get $\mathbf{1+a=0\Leftrightarrow a=-1}$ and $\mathbf{1-a^2=0\Leftrightarrow a=\pm 1}$}$\\\\ $\boxed{\boxed{\mathbf{a=-1}}}$

x²(a+1)+x.(1-a²)=0
How can it be this 0.x²+0.x?Are those both the positive numbers that you are applying this rule?
=>x[x(1+a)+1-a²]=0
=>x=0 or x(1+a)+1-a²=0
x(1+a)+1-a²=0
=>x=a²-11+a=a-1
If x is known then a can be calculated.But there is no x.