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Let $mathbbQ^+$ denote the set of positive rational numbers. Let $f : mathbbQ^+ to mathbbQ^+$ be a function such that
$f left( x + fracyx right) = f(x) + fracf(y)f(x) + 2y$ for all $x,$ $y in mathbbQ^+.$

Find all possible values of $f left( frac13 right).$

If I substitute in $x=y$, then I get $f(x+1)-f(x)=2x+1$. This suggests that $f(x)=x^2$ works, and one possible value of $f(1/3)$ is $1/9$. Did I miss anything?

Yes.

But what you didn't do was prove that you had all possible values. If somebody told you to solve $x^2=1$, you wouldn't say $1$ and leave it. You would prove that $1$ is the only solution, or find more solutions and prove that you'd found all of them.

A similar things applies here. The question asks for you to find all possible values. How do you know you have all of them?

Here's a quick sketch of how to prove that you have all solutions:
Set $y:=y+x$ and subtract the original equation to get
$$2x+2fracyx+1=fracf(y+x)-f(y)f(x)+2x$$Rearrange and simplify:

$$2fracyxf(x)+f(x)=f(y+x)-f(y)$$
Repeat the same trick: set $y:=y+1$ and subtract that last equation to get
$$2fracf(x)x=2y+2x+1-2y-1$$
so $f(x)=x^2$ as required. Just verify this works by substituting back into the original equation.

From $f(x+1)-f(x)=2x+1$, we can deduce that $f(x)=x^2+f(1)-1$ for all $xinmathbbZ^+$.

Note that we also have $f(1+frac x1)=f(1)+fracf(x)f(1)+2x$.

Therefore, $f(x+1)=f(x)+2x+1=dfracf(x)f(1)+2x+f(1)$.

Hence, $f(x)left(1-dfrac1f(1)right)=f(1)-1$ for $xinmathbbQ^+$.

As $f$ is not constantly zero, $f(1)=1$. So, $f(x)=x^2$ for $xinmathbbZ^+$.

$f(3+frac13)=f(3)+fracf(1)f(3)+2(1)=3^2+frac13^2+2=11+frac19$

$f(2+frac13)=f(3+frac13)-2(2+frac13)-1=5+frac49$

$f(1+frac13)=f(2+frac13)-2(1+frac13)-1=1+frac79$

$f(frac13)=f(1+frac13)-2(frac13)-1=frac19$

We have
$$tag$x=y=1$ f(2)=f(1)+3$$
$$tag$x=1,y=2$ f(3)=f(1)+fracf(2)f(1)+4=f(1)+5+frac3f(1)$$
$$tag$x=y=2$ f(3)=f(2)+5=f(1)+8$$
hence $f(1)=1$.
Let $S=,xinBbb Q_+mid f(x)=x^2,$.
As just seen, $1in S$.
From the functional equation we see that if $x$ and one of $y,x+frac yx$ are $in S$, then so is the third. In other words,
$$tag1x,yin Simplies x+frac yxin S $$
and
$$tag2 x,yin Sland x<yimplies (y-x)xin S.$$
In particular, using $(1)$ with $y=x$ and $(2)$ with $x=1$, we find that for $xin Bbb Q_+$,
$$tag3xin Siff x+1in S$$ and so by induction $Bbb Z_+subseteq S$.

Let $x=frac abinBbb Q_+$. From $(1)$, $b+xin S$. Then by applying $(3)$ $b$ times, $xin S$. In other words, $S=Bbb Q_+$ and
$$f(x)=x^2$$
for all $xinBbb Q_+$.