If $\;f(x)+\frac{1}{x}=f^{-1}(\frac{1}{f(x)})\;$ and $f^{-1}(\frac{1}{f(x)})>0$. Then $f(x)$ is? If $f'(x)>0\; \forall \;x\in \mathbb R^+\;$ where $f:\mathbb R^+\rightarrow \mathbb R\;$ and $\;f(x)+\dfrac{1}{x}=f^{-1}\bigg(\dfrac{1}{f(x)}\bigg)\;$ and $f^{-1}\bigg(\dfrac{1}{f(x)}\bigg)>0$. Then $f(x)$ is?.
My Approach:
I replaced $f(x)$ by $\dfrac{1}{f(x)}\implies x\rightarrow f^{-1}\bigg(\dfrac{1}{f(x)}\bigg)$.
$\implies \; \dfrac{1}{f(x)}+\dfrac{1}{f^{-1}\bigg(\dfrac{1}{f(x)}\bigg)}=f^{-1}(f(x))$
$\implies\;$ $\dfrac{1}{f(x)}+\dfrac{x}{xf(x)+1}=x$
$\implies\;$$ \dfrac{2xf(x)+1}{f(x)(xf(x)+1)}=x\; \implies \{xf(x)\}^2-\{xf(x)\}-1=0$
$\implies\;$ $xf(x)=\dfrac{1-\sqrt{5}}{2}$ and $xf(x)=\dfrac{1+\sqrt{5}}{2}$.
But answer given is $f(x)=\dfrac{\sqrt{5}-1}{2x}$.
My Doubts:
$1.$ What is wrong in my method?
$2.$ Did I make any mistake by putting $f(x)\rightarrow \dfrac{1}{f(x)}?$
 A: The given answer is wrong, as can be verified in WA. OP's results are correct (also in WA), but only the negative solution satisfies the monotonicity condition $\,f'(x) \gt 0\,$ on $\,\mathbb R^+\,$.
Below is an alternate proof with the steps and substitutions made more explicit.

$f(x)+\dfrac{1}{x}=f^{-1}\left(\dfrac{1}{f(x)}\right)$

Let $\,y = f(x)\,$, then $\,x = f^{-1}(y) = h(y)\,$ and the equation becomes $\,y + \dfrac{1}{h(y)}=h\left(\dfrac{1}{y}\right)\,$, or:
$$
y \, h(y) + 1 = h(y)\,h\left(\dfrac{1}{y}\right) \tag{1}
$$
Writing $\,(1)\,$ for $\,y \mapsto \dfrac{1}{y}\,$:
$$
\dfrac{1}{y} \,h\left(\dfrac{1}{y}\right) + 1 = h\left(\dfrac{1}{y}\right)\,h(y) \;\;\iff\;\; h\left(\dfrac{1}{y}\right)\,\big(y\,h(y) - 1\big) = y \tag{2}
$$
Multiplying $\,(1) \times(2)\,$ and canceling out $\,h\left(\dfrac{1}{y}\right) \gt 0\,$:
$$
\require{cancel}
\big(y \, h(y) + 1\big)\,\big(y \, h(y) - 1\big)\,\cancel{h\left(\dfrac{1}{y}\right)} = y\,h(y)\,\cancel{h\left(\dfrac{1}{y}\right)}
\\ \iff\quad y^2\,h^2(y) - y\,h(y) - 1 = 0
 \quad\iff\quad h(y) = \dfrac{1 \pm \sqrt{5}}{2y}
$$
It is easily verified that $\,h\big(h(y)\big) = y\,$, so $\,f(x)=f^{-1}(x)=h(x)=\dfrac{1 \pm \sqrt{5}}{2x}\,$.
Both solutions satisfy the additional constraint $\,f^{-1}\left(\dfrac{1}{f(x)}\right) \gt 0\,$ for $\,x \in \mathbb R^+\,$, because $\,f\,$ is of the form $\,f(x)=f^{-1}(x) = \dfrac{\lambda}{x}\,$, and therefore $\,f^{-1}\left(\dfrac{1}{f(x)}\right) = \dfrac{\lambda^2}{x} \gt 0\,$ when $\,x \gt 0\,$.
Only the negative solution satisfies the condition $\,f'(x) \gt 0\,$, so in the end $\,f(x) = \dfrac{1-\sqrt{5}}{2x}\,$.
