We have $f(1;kt;t^2(k+1))=t(k+1)\Rightarrow f(1;a;b)=\dfrac{a+\sqrt{a^2+4b}}{2}$, for all $a,b\in\Bbb{S}$. Now $f(b;a;1)= \dfrac{a+\sqrt{a^2+4b}}{2b}\Rightarrow f(x;y;z)=f(x;\dfrac{y}{\sqrt{z}}*\sqrt{z};1*(\sqrt{z})^2)=\sqrt{z}*f(x;\dfrac{y}{\sqrt{z}};1)= \dfrac{y+\sqrt{y^2+4xz}}{2x}$ , for $x,y,z\in\Bbb{S}$.

That's a really nice and fresh olympiad problem!

In general, the solution to the first 2 equations seems to be $f(x,y,z)=h(y^2/xz)\sqrt{z/x}$,where h is a one variable function. Proof: Let $g(x,y,z)=xf(x,y,z)$. The first equations says that g is symmetrical in x and z. Next let $k(x,c,z)=g(x,c\sqrt{xz}, z)$. The second equation says that for fixed x and c, $k$ is proportional to $\sqrt{z}$. Then, letting x vary too, it must be proportional to $\sqrt{xz}$ by the first equation's symmetry. So this gives us the result that f is the product of $\sqrt{z/x}$ and a function only dependent on c, or y^2/xz as claimed. This does the problem as k^2/(k+1) takes all positive values uniquely as k varies across the positive reals.

Note that if $x=k^2x_1, y=ky_1, z=z_1$, then \[kx_1f(k^2x_1, k^2y_1, k^2z_1)=kx_1f(x, ky, k^2z)=k^2x_1 f(x,y,z)=\] \[k^2x_1f(k^2x_1, ky_1, z_1)=z_1f(z_1, ky_1, k^2x_1)=kz_1f(z_1, y_1, x_1)=kx_1f(x_1, y_1, z_1).\] Since the domain of the function $f$ is the set of positive real numbers, we get $f(ta, tb, tc) =f(a,b,c)$, for any $t > 0$. Therefore $f(a,b,c)= f\left (1, \frac{b}{a}, \frac{c}{a}\right)$. Let $kl= \frac{b}{a}$ and $k^2(l+1)=\frac{c}{a}$. Then $k\frac{b}{a}+k^2=\frac{c}{a}$. Solving quadratic equation, we get $k= \frac{-b \pm \sqrt{b^2 +4ac}}{2a}$. Hence \[f(a,b,c)= f(1, \frac{b}{a}, \frac{c}{a})= f(1, kl, k^2 (l+1))=kf(1,l, l+1))=k(l+1)\] \[f(a,b,c)=k^2(l+1) \cdot \frac{1}{k}= \frac{c}{a} \cdot \frac{2a}{-b +\sqrt{b^2 +4ac}},\] where the sign in the expression for $k$ is a "+", because $f$'s range is the set of positive real numbers. Answer: $f(a,b,c)= \frac{2c}{-b+\sqrt{b^2 +4ac}}=\frac{\sqrt{b^2 +4ac}+b}{2a}$. Finally, we need to check that this function satisfies all the given conditions, and this is true.

Hmmm... the answer to this problem is the solution $t$ of $xt^2-yt-z=0$. Is there a nice, intuitive solution using this? I just bashed this by forcing the third term to become one then reversing it then forcing the difference between the 2nd and 3rd terms to become 1.

@above not sure how "intuitive" of an answer this is, but i think the very rigid conditions basically already show to you that the function is basically uniquely defined from what they give you through just algebra and scaling and stuff, which makes u go for a more "equation solving" approach as opposed to actually dealing with functional equations

The answer is $f(x, y, z) = \frac{y + \sqrt{y^{2} + 4xz}}{2x}$, which can easily be checked that this works. First we calculate $f(1, y, z)$. Consider some $r$ such that $yr + 1 =zr^{2}$, then from condition two we have $f(1, y, z) = \frac{f(1, yr, zr^{2})}{r} = zr$. From the quadratic formula, we have \[r = \frac{y + \sqrt{y^{2} + 4z}}{2z} \Rightarrow f(1, y, z) = \frac{y + \sqrt{y^{2} + 4z}}{2}\]We can also use condition one to get \[f(z, y, 1) = \frac{1}{z}f(1, y, z) = \frac{y + \sqrt{y^{2} + 4z}}{2z}\] Then, we have \[f(x, y, z) = \sqrt{z}f(x, \frac{y}{\sqrt{z}}, 1) = \sqrt{z}\left(\frac{y}{\sqrt{z}} + \sqrt{\frac{y^{2}}{z} + 4x}{2x}\right) = \frac{y + \sqrt{y^{2} + 4xz}}{2x}\]

Let $P(x,y,z,k)$ be the second assertion. Take $P\left(x,\frac{y}{\sqrt{z}},1,\sqrt{z}\right)$: \[f(x,y,z)=\sqrt{z}\cdot f\left(x,\frac{y}{\sqrt{z}}, 1\right)\overset{(a)}{=}\frac{\sqrt{z}}{x}\cdot f\left(1,\frac{y}{\sqrt{z}}, x\right)\]Now, let $t$ be the positive solution of the equation \[ t^2x=t\cdot \frac{y}{\sqrt{z}}+1\]Clearly $t=\frac{\frac{y}{\sqrt{z}}+\sqrt{\frac{y^2}{z}+4x}}{2x}$. Anyways, take $P(1,\frac{y}{\sqrt{z}},x,t)$: \[t\cdot f\left(1,\frac{y}{\sqrt{z}},x\right)=f\left(1,t\cdot \frac{y}{\sqrt{z}},t\cdot \frac{y}{\sqrt{z}} +1\right)\overset{(c)}{=} t\cdot \frac{y}{\sqrt{z}} +1=t^2x\]Finally \[f(x,y,z)=\frac{\sqrt{z}}{x}\cdot tx=\frac{y+\sqrt{y^2+4xz}}{2x}\]

We begin by plugging in $(x,y,z,k) = \left(1,y,z,\frac{y+\sqrt{y^2+4z}}{2z}\right)$ in (b) where that value of $k$ is chosen so that $k^2z - ky = 1$ in order to use (c): \[ky+1=f(1,ky,k^2z)=kf(1,y,z)\Longrightarrow f(1,y,z) = y+\frac{1}{k} = y+\frac{2z}{y+\sqrt{y^2+4z}}\quad \forall y,z\in\mathbb R^+\]Now plugging $(x,y,z)=(1,y,z)$ into (a) gives \[y+\frac{2z}{y+\sqrt{y^2+4z}} = f(1,y,z) = zf(z,y,1)\Longrightarrow f(z,y,1) = \frac{y}{z}+\frac{2}{y+\sqrt{y^2+4z}}\]Finally, plugging $(x,y,z,k)=\left(x,y,z,\frac{1}{\sqrt{z}}\right)$ into (b) gives: \[f(x,y,z) = \frac{1}{k}f\left(x,\frac{y}{\sqrt{z}},1\right) = \sqrt{z}\left(\frac{y}{x\sqrt{z}}+\frac{2}{\frac{y}{\sqrt{z}}+\sqrt{\frac{y^2}{z}+4x}}\right) = \frac{y}{x}+\frac{2z}{y+\sqrt{y^2+4xz}}\]Now we check that this function does indeed satisfy all 3 conditions: \[\text{For (a): }xf(x,y,z) = y+\frac{2xz}{y+\sqrt{y^2+4xz}} = zf(z,y,x)\]\[\text{For (b): }f(x,ky,k^2z) = \frac{ky}{x}+\frac{2k^2z}{ky+\sqrt{k^2y^2+4k^2xz}} = \frac{ky}{x}+\frac{2kz}{y+\sqrt{y^2+4xz}} = kf(x,y,z)\]\[\text{For (c): }f(1,k,k+1) = \frac{k}{1}+\frac{2(k+1)}{k+\sqrt{k^2+4(k+1)}}=k+1\]Therefore the only solution is $\boxed{f(x,y,z) = \frac{y}{x}+\frac{2z}{y+\sqrt{y^2+4xz}}}$.

The only solution is $f(x,y,z) = \frac{y + \sqrt{y^2 + 4xz}}{2x}$. We may check that this works. Notice that $f(1,ky,k^2z) = kf(1,y,z)$. Setting \[k = \frac{y + \sqrt{y^2 + 4z}}{2},\]we have $k^2z = ky + 1$, so \[kf(1,y,z) = k^2 z\implies f(1,y,z) = z\cdot \frac{y + \sqrt{y^2 + 4z}}{2z} \] Now, $f(1,y,z) = zf(z,y,1)$, so $f(z,y,1) = \frac{y + \sqrt{y^2 + 4z}}{2z}$. In the second equation, set $k = \frac{1}{\sqrt{z}}$. We get \[f\left(x, \frac{y}{\sqrt{z}}, 1\right) = \frac{f(x,y,z)}{\sqrt{z}},\]so \[f(x,y,z) = \sqrt{z} \cdot f\left(x, \frac{y}{\sqrt{z}}, 1\right) = \frac{y + \sqrt{y^2 + 4xz}}{2x},\]as desired.

The unique answer is $f(a, b, c) \equiv \frac{b}{2a} + \frac{1}{2a}\sqrt{b^2 + 4ac}$, i.e. the positive solution to $ax^2 - bx - c = 0$. It is easy to check that such $f$ must satisfy the three conditions presented. Let $b, c \in \mathbb{R}^+$ be arbitrary. Let $k_0 = \frac{b + \sqrt{b^2 + 4c}}{2c}$ be the positive solution to $k_0^2c = k_0b + 1$. From the conditions, we obtain: \[f(1, b, c) = \frac{1}{k_0}f(1, k_0b, k_0^2c) = \frac{1}{k_0}f(1, k_0^2c - 1, k_0^2c) = k_0c =\frac{1}{2}b + \frac{1}{2}\sqrt{b^2 + 4c}.\] To finish, note that the conditions require: \[f(a, b, c) = \sqrt{c}f\left(a, \frac{b}{\sqrt{c}}, 1\right) = \frac{\sqrt{c}}{a}f\left(1, \frac{b}{\sqrt{c}}, a\right) = \frac{\sqrt{c}}{a}\left(\frac{1}{2}\frac{b}{\sqrt{c}} + \frac{1}{2}\sqrt{\frac{b^2}{c} + 4a}\right) = \frac{b}{2a} + \frac{1}{2a}\sqrt{b^2 + 4ac}\]for all $(a, b, c) \in \mathbb{R}^3$.

We try to find $f(x, y, z)$. We can see that \begin{align} f(x, y, z) &= \sqrt{z} f\left(x, \frac{y}{\sqrt{z}}, 1\right) \\ xf\left(x, \frac{y}{\sqrt{z}}, 1\right) &= f\left(1, \frac{y}{\sqrt{z}}, x \right) \\ k f \left(1, \frac{y}{\sqrt{z}}, x \right) &= kx^2 \end{align} Where $k$ is the solution to $xk^2 - \frac{ky}{\sqrt{z}} = 1$. We can use the quadratic formula to find that \[k = \frac{\frac{y}{\sqrt{z}} \pm \sqrt{\frac{y^2}{z} + 4x}}{2x}\]We can now back track through our conditions to find that $f(x, y, z) = \boxed{\frac{y \pm \sqrt{y^2 + 4xz}}{2x}}$. $\blacksquare$

Consider the following claim: Claim: $f(1, a, b) = a + \frac{2b}{a + \sqrt{a^2 + 4b}}$. Proof. Note that $f(1, yk, (y+1)k^2) = kf(1, y, y+1) = (y+1)$. Now we'll find $y, k$ such that $a = yk$ and $b = (y+1)k^2$. Finding $k$, we get $b = (y+1)\frac{a^2}{y^2}$ and solving it, we get $y = \frac{a^2 + a\sqrt{a^2 + 4b}}{2b}$, so $k = \frac{2b}{a + \sqrt{a^2 + 4b}}$. Hence $f(1, a, b) = f(1, yk, (y+1)k^2) = k(y+1) = a + \frac{2b}{a + \sqrt{a^2 + 4b}}$, as needed. $\blacksquare$ Using claim, we get $bf(b, a, 1) = f(1, a, b) = a + \frac{2b}{a + \sqrt{a^2 + 4b}}$, so $f(b, a, 1) = \frac{a}{b} + \frac{2}{a + \sqrt{a^2 + 4b}}$. Now we'll find $f(x, y, z)$. Note that $f(x, y, z) = \sqrt{z}f(x, \frac{y}{\sqrt{z}}, 1) = \sqrt{z} \cdot (\frac{y}{\sqrt{z}x} + \frac{2}{\frac{y}{\sqrt{z}} + \sqrt{\frac{y^2}{z} + 4x}}) = \frac{y}{x} + \frac{2z}{y + \sqrt{y + 4xz}}$. Thus we're done. $\blacksquare$

Solution:

This problem is hilarious. I claim that $f(x, y, z)$ gives the unique positive root to the equation $xt^2-yt-z=0$, which obviously satisfies the given properties. Let $g(x, y, z)$ denote this function; we aim to show that $f \equiv g$. Now, we define a move to be a transformation of coefficients $(x, y, z) \to (z, y, x)$ or $\left(x, ky, k^2 z\right) \to (x, y, z)$. It suffices to show that we can get from $(x, y, z) \to (1, k, k+1)$ for some real number $k$ in a finite number of moves. Indeed, if $f(x, y, z) = g(x, y, z)$, then for $(x', y', z')$ the preimage of $(x, y, z)$ under a move, we have $f(x', y', z') = g(x', y', z')$ too. This is quite doable. Indeed, we let $k$ be a small real number and consider \[(x, y, z) \to \left(x, \frac yk, \frac z{k^2}\right) \to \left(\frac z{k^2}, \frac yk, x\right) \to \left(\frac z{k^2}, \frac y{k \sqrt x}, 1\right) \to \left(1, \frac y{k \sqrt x}, \frac z{k^2}\right).\]By letting $k$ be small, we can force $\frac y{k\sqrt x} < \frac z{k^2}$, and now scale the second and third entries down by $c$ and $c^2$ for some $c$ so that they read $(1, \ell, \ell + 1)$ for some $\ell$, as needed. This completes the proof.