find function with condition using Leibniz okay, so we have $f:(0, +\infty) \to (0, +\infty)$, with $f(1)=2$, such as $f'(\frac{1}{x})=\frac{1}{f(x)}$.
Now, I know we're supposed to use Leibniz's binomial theorem. And the answer to the problem is $$f(x)=(1+x)e^\frac{1-x}{2+2x}$$
Further than that, I've tried working with the condition but i'm getting nowhere everytime and the answer in the textbook has no other suggestions. Any help is welcome at this point.

Hint:
If
$$
f'\left(\frac{1}{x}\right)=\frac{1}{f(x)}
$$
then
$$
f'(x)=\frac{1}{f\left(\frac{1}{x}\right)}
$$
and
\begin{align}
\frac{d}{dx} \left(
f(x)f\left(\frac{1}{x}\right)\right)&=f'(x) f\left(\frac{1}{x}\right)-\frac{f(x)f'\left(\frac{1}{x}\right)}{x^2}\\
&=1-\frac{1}{x^2}\ .
\end{align}
Therefore, by integrating this equation from $1$ to $\ x\ $, we get
\begin{align}
f(x) f\left(\frac{1}{x}\right)-f(1)^2&= f(x) f\left(\frac{1}{x}\right)-4\\
&=x+\frac{1}{x}-2\ ,\\
\end{align}
—that is,
\begin{align}x+\frac{1}{x}+2&=f(x) f\left(\frac{1}{x}\right)\\
&=\frac{f(x)}{f'(x)}\ ,\\
\end{align}
or
\begin{align}
\frac{x}{(1+x)^2}&= \frac{f'(x)}{f(x)}\\
&=\frac{d}{dx}\ln f(x)\ .
\end{align}
Can you finish it off from here?
