Should the second condition perhaps be $f\left({1\over x}\right)={f(x)\over x^{\color{red}{2}}}$?

Farenhajt wrote: Should the second condition perhaps be $f\left({1\over x}\right)={f(x)\over x^{\color{red}{2}}}$? No, it is $x^3.$

A sketch: Show that using the Euclidean algorithm we can express any rational number by starting from 1 and doing the transformations $x \to 1/x$ and $x \to \frac{x}{x+1}$. Thus if we specify $f(1)$ the function is uniquely defined. Let $f(1) = c$ for a positive rational $c$, and notice that $f(p/q) = cp^2/q$ satisfies both equations, so this is all the solutions.

Let $g(x)=x^2f\left(\frac1x\right)$, then we get an functional equation on $g:\mathbb Q^+\to\mathbb Q^+$ like as following: $g\left(x\right)=xg\left(\frac1x\right)\qquad(1)$ and $\left(x+1\right)g\left(\frac1x\right)=g\left(\frac{x+1}x\right)\qquad(2)$. These imply $g\left(x\right)=g\left(\frac x{x+1}\right)$ for all $x \in \mathbb{Q^+}$. Take $x=\frac1n$, we get $g\left(\frac1n\right)=c$ for all $n\in\mathbb Z^+$. Here we denote the constant $c=f(1)$. Since (1), we have $\left(1+\frac1x\right)g\left(\frac x{x+1}\right)=g\left(\frac{x+1}x\right)$, take $x=\frac1n$, then $g(1+n)=(n+1)g\left(\frac1{n+1}\right)=c(n+1)$. This implies $g(n)=cn$ for all $n\in\mathbb Z^+$. Now we have, \[g\left(n+x\right)=g\left(1+n-1+x\right)=\left(n+x\right)g\left(\frac1{n-1+x}\right)=\frac{n+x}{n-1+x}g\left(n-1+x\right).\] This shows, $\frac{g\left(n+x\right)}{n+x}=\frac{g\left(x\right)}{x}$ for all $x\in\mathbb Q^+$. Now we apply Euclidean Algorithm: for each $r=\frac mn\in\mathbb Q^+$, we write $m=nq_1+r_1$, $n=r_1q_2+r_2$, $\ldots$, $r_k=r_{k+1} q_{k+2}$. Then $r_{k+1}=\gcd\left(m,n\right)$. So, \[ g\left(r\right)=g\left(q_1+\frac{r_1}n \right)=\frac{m}{n}g\left(\frac n{r_1}\right)=\cdots= \] \[ =\frac mn\cdot \frac {n}{r_1}\cdots \frac{r_{k-1}}{r_{k}}g\left(\frac{r_k}{r_{k+1}}\right)=c\frac{m}{r_{k+1}}=c\frac{m}{\gcd(m,n)}. \] This shows that $g\left(\frac mn\right)=c\frac{m}{\gcd(m,n)}$, where $c$ is a constant. So, $f\left(\frac mn\right)=\left(\frac mn\right)^2g\left(\frac nm\right)=c\frac{m^2}{n\cdot\gcd(m,n)}$ for any $m,n$ are positive integers. In other form, if $r=\frac mn$, where $\gcd(m,n)=1$, then $f(r)=c\cdot\frac{m^2}n$, as MellowMelon show.

Multiplying both conditions by a constant a and considering the new function g(x)=af(x) leave conditions unchanged, hence we can suppose WLOG f(1)=1. Replace x by $\frac{1}{x}$ in first condition and use second to get : $f(x+1) = (\frac{x+1}{x})^2 f(x)$, and by immediate induction : $f(x+n)=(\frac{x+n}{x})^2 f(x)$ (*). In particular, letting x=1, we get for all positive integers : f(n)=n². Now, let a be a rational number and write $a=[a_1,a_2,...,a_n]$ (finite continued fraction). Using (*) and then the second condition, we get : $f(a)=f([a_1,a_2,...,a_n])=([a_1,a_2,...,a_n])^2([a_1,a_2,...,a_n]-a_1)f([a_2,...,a_n])$. By straight induction, it follows that : $f(a) = (\prod_{i=1}^{n-1}[a_i,a_{i+1},...,a_n]^2([a_i,a_{i+1},...,a_n]-a_i)) a_n^2$, which completely determines the function. Conversely, any function which may be written in the preceding form times some constant verifies both condition 2 (notice that if $a=[a_1,a_2,...,a_n]>1$, then $\frac{1}{a}=[0,a_1,a_2,...,a_n]<1$), and condition 1 ( now equivalent to $f(x+1) = (\frac{x+1}{x})^2 f(x)$, which may now be proven easily).

Cool Solution!