<s>[INST] Composite Function value Given: $f(f(x)).[1+f(x)] = -f(x)$. Find $f(3)$
Approach to the question:-
$fof(x) + f(x).fof(x) = -f(x)$
$fof(x) + f(x) + f(x).fof(x)+1=1$
$(1+f(x))(1+f(f(x))=1$
This implies that $1+f(x)$ and $1+fof(x)$ are reciprocals.
This gives $f(x)=\frac{1}{x}-1$ or $ f(x)=\frac{-1}{x}-1$ as solutions. (By inspection)
Therefore, $f(3)=\frac{-2}{3}$ or $f(3)=\frac{-4}{3}$ 
But the answer given is $f(3)=\frac{-3}{4}$
What is my mistake?
 [/INST] Well $1+f(x)$ and $1+f(f(x))$ are reciprocals of each other and not of $x$ which means $$1+f(x)=\frac1{1+f(f(x))}\\1+f(f(x))=\frac{1}{1+f(x)}$$
Consider this instead since $f(x)=-1$ doesn't satisfy the above condition we can divide by $1+f(x)$ to get
$$f(f(x))=\frac{-f(x)}{1+f(x)}$$
Now putting $f(x)=t$ (this step requires justification like that f is onto though I guess the problem assumes it) we get
$$f(t)=\frac{-t}{1+t}$$
So $f(3)=\frac{-3}{4}$
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