$f(x)$ such that $f(x)*f(f(x)+\frac{1}{x})=1$ and $f(x) > - \frac{1}{x}$ $ \forall$ $ x > 0$ and $f(x)$ is an injective function I don't even know where to begin in solving this functional equation. I got this in a multiple choice question exam and was able to solve it by substituting all of the given options into the equation. The one that worked is $$f(x) = \frac{1 -\sqrt 5}{2x}$$
I can't even prove that this is a unique solution. However, I was able to show that $f(1)=\frac{1-\sqrt 5}{2}$ 

Let's for a moment assume that $f(x)$ is invertible with $g(x)$ being its inverse. We can always discard our assumption later on if things don't turn out the way we want them to.
So, $f(x)f\biggl(f(x) + \frac{1}{x}\biggr) = 1$ can be modified as $x.f\biggl(x + \frac{1}{g(x)}\biggr) = 1$
$$\Rightarrow x + \frac{1}{g(x)} = g(\frac{1}{x})$$
$$\Rightarrow   g(\frac{1}{x}) - \frac{1}{g(x)} = x$$
Substituting $x$ for $\frac{1}{x}$, we get $ g({x}) - \frac{1}{g(\frac{1}{x})} = \frac{1}{x}$
Eliminating $g(\frac{1}{x})$ from the above equations, we get 
$$g(x)\bigl(g(x)-\frac{1}{x}\bigr) = \frac{1}{x^2}$$
$$\Rightarrow x^2g^2(x) - xg(x) -1 =0$$
This can be modified as
$$f^2(x)x^2 - f(x)x - 1=0$$
which is simply a quadratic equation leading to the solution:
$$f(x) = \frac{1 \pm \sqrt{5}}{2x}$$
Now, to decide between the two possible solutions, one could substitute and check for $f(x)$ in the question. Both $f(x) = \frac{1 \pm \sqrt{5}}{2x}$ satisfy the conditions of the question. Also, our original assumption remains true that $f(x)$ is an invertible function, hence no contradictions here.
