Find the value of $\alpha$ Let $f(x)=\frac{\alpha x}{x+1},x\neq-1$. Then for what value of $\alpha$ is $f(f(x))=x$?

My approach:
Given that, 
 $f(f(x))=x$
Applying $f^{-1}$ on both sides, we get
$f(x)=f^{-1}(x)$
The above condition is only true when the curve $y=f(x)$ intersects with $y=x$.

$\therefore f(x)=x\implies \frac{\alpha x}{x+1}=x$
 On solving this, we get $\alpha $ as a function of $x$.

I know that there is another method to solve this in which we find the expression for $f(f(x))$ and equate it to $x$.

I don't need that solution. I just want to know what is wrong with this approach as this was the first thing that came to my mind on seeing the question and the method was much shorter compared to the above-mentioned method.

Anybody with alternative methods are welcome to share.

Thanks in advance.
Note that when $\alpha = -1$, the vertical and horizontal asymptotes will be equal: $x=-1, y=-1$ and $f(x)=f^{-1}(x)$. 

In fact, that is the case for any hyperbola with equal asymptotes:

$$f(x)=\frac{ax+b}{c x+d}, \left(-\frac{d}{c}=\frac{a}{c}\right) \Rightarrow \\
y=\frac{ax+b}{c x+d} \Rightarrow x=\frac{b-yd}{yc-a} \Rightarrow \\
f^{-1}(x)=\frac{b-xd}{xc-a}=\frac{ax+b}{cx+d}.$$
You can equate the original function to its inverse to see the solution $\alpha=-1$ clearly:
$$f(x)=f^{-1}(x) \Rightarrow \frac{\alpha x}{x+1}=\frac{x}{\alpha-x} \Rightarrow \\
\frac{\alpha^2x-\alpha x^2-x^2-x}{(x+1)(\alpha -x)}=0 \Rightarrow \\
\frac{(\alpha +1)(\alpha-x-1)x}{(x+1)(\alpha -x)}=0$$
$f(f(x))=\alpha^2x/((x+1)(\alpha x/(x+1)+1))$
The equation you have to solve is:
$$f(f(x))=x$$
So you get:
$$\alpha=-1$$
and
$$\alpha=x+1$$