Sketch: suppose $f(1)=c, f(2)=c+d$, then note $S=\{ x: f(x)=c+d(x-1)\}$ is dense in $\mathbb{R}^+$. So if we can find $f(x)<c+d(x-1)$, then we pick $z$ small, $y^2+z^2=x(y+z)$, $f(y)=c$, then $yf(y)+zf(z)=(z+y)f(x)$. We pick $z$ such that $f(z)<0$.

One can check that non-decreasing, linear functions work, so we only have to prove that $f$ must be non-decreasing and linear, i.e., $f = x \mapsto ax + b$ for some $a, b \geq 0$ with either $a > 0$ or $b > 0$. Looking at the form of the equation motivates the following solution. Define a continuous function $g : (\mathbb{R}^+)^2 \rightarrow \mathbb{R}^+$ by $g(x, y) = \frac{x^2 + y^2}{x + y}$ for all $x, y \in \mathbb{R}^+$. Now, fix $x$ and $y$, let $z = g(x, y)$, and consider the functional equation: \[ xf(x) + yf(y) = (x + y)f(z). \]Manipulating gives: \begin{align*} y f(y) - y f(x) &= (x + y) f(z) - (x + y) f(x) \\ \frac{f(y) - f(x)}{y - x} &= (f(z) - f(x)) \frac{x + y}{y(y - x)} \end{align*}However, note that $\frac{y(y - x)}{x + y} = \frac{y^2 - xy}{x + y} = \frac{x^2 + y^2}{x + y} - x = z - x$. So, \[ \frac{f(y) - f(x)}{y - x} = \frac{f(z) - f(x)}{z - x}. \]Hence, in Cartesian coordinate, the points $(x, f(x))$, $(y, f(y))$, and $(z, f(z))$ are collinear. For simplicity, for each $x \in \mathbb{R}^+$, we denote the point $S_x = (x, f(x))$, which always lies in the quadrant 1. We conclude that $S_x, S_y, S_{g(x, y)}$ are always collinear for each $x, y \in \mathbb{R}^+$. Next, for each $0 < x < y$, we study the property of the set $T(x, y) = \{x, y\} \cup \{z \in \mathbb{R}^+ | S_x, S_y, S_z \text{ are collinear } \}$. It is easy to check that for each $z_1, z_2 \in T(x, y)$, we have $g(z_1, z_2) \in T(x, y)$. Claim 1: $T(x, y) \cap [x, y]$ is dense in $[x, y]$. Proof: Suppose not; then there exists $z \in [x, y]$ and $\varepsilon > 0$ such that the interval $(z - \varepsilon, z + \varepsilon)$ does not intersect with $T(x, y)$. Since $x, y \in T(x, y)$, we have $z \in (x, y)$ and $(z - \varepsilon, z + \varepsilon) \subseteq (x, y)$. Take the longest such open interval containing $z$ that doesnot intersect $T(x, y)$, say $I = (a, b)$, where $x \leq a < z < b \leq y$. Then, for any $\delta > 0$, there exists $a', b' \in T(x, y)$ with $a - \delta < a' \leq a$ and $b + \delta > b' \geq b$. Now, we claim that if $\delta$ is small enough, any pick of $a'$ and $b'$ satisfies $g(a', b') \in I$, which is a contradiction since $g(a', b') \in T(x, y)$. By AM-GM, whenever $\delta < b$, for any choice of $a'$ and $b'$, \[ g(a', b') \geq \frac{a' + b'}{2} > \frac{a + b - \delta}{2} > a. \]On the other hand, \[ g(a', b') = b' - \frac{b' - a'}{b' + a'} a' < b + \delta - \frac{b - a}{b + a + \delta} (a - \delta). \]If $\delta$ is small enough, then we can guarantee that \[ \delta < \frac{b - a}{b + a + \delta} (a - \delta) \iff \frac{a - \delta}{\delta} > \frac{b + a + \delta}{b - a} \iff \frac{a}{\delta} > \frac{2b + \delta}{b - a}. \]The last inequality is certainly true for small $\delta$ since the LHS goes to $\infty$ while the RHS stays finite. Thus, for $\delta$ small enough, $g(a', b') < b$. Hence, $a < g(a', b') < b$, and thus $g(a', b') \in I$, a contradiction. We conclude that $T(x, y) \cap [x, y]$ is dense in $[x, y]$. Claim 2: For any $0 < x < y$, $T(x, y)$ contains arbitrarily large elements. Proof: Let $y = x_1$, and iteratively for each $n \geq 2$, we set a positive real number $x_n > x_{n - 1}$ such that $x_{n - 1} = \frac{x^2 + x_n^2}{x + x_n}$. Equivalently, \[ x_n^2 - x_{n - 1} x_n + x^2 - x x_{n - 1} = 0, \]\[ x_n = \frac{x_{n - 1} + \sqrt{x_{n - 1}^2 + 4 x (x_{n - 1} - x)}}{2} \]Since $x_1 > x$, we can easily prove by induction that $x_n > x_{n - 1} > x$ for all $n \geq 1$. In particular, for all $n \geq 2$, \[ \frac{x_n}{x_{n - 1}} = \frac{1 + \sqrt{1 + 4 x (1 - x x_{n - 1}^{-1})}}{2} \geq \frac{1 + \sqrt{1 + 4 x (1- xy^{-1})}}{2}. \]The right hand side is constant with respect to the sequence and is greater than $1$, so the term grows at least exponentially fast. Thus, it cannot be bounded above. On the other hand, we can also check by induction that $x_n \in T(x, y)$ for each $n \geq 1$, so we are done. These two lemmas prepare us for the final attack. Claim 3: $f$ is non-decreasing and linear. Proof: First, we prove that $f$ is non-decreasing. Suppose not; then $f(x) > f(y)$ for some $x < y$. Let $m = \frac{f(y) - f(x)}{y - x}$ and $c = f(x) - mx$. Then, $f(z) = mz + c$ for all $z \in T(x, y)$. Since $f(x) > f(y)$ and $x < y$, we have $m < 0$. Then, for arbitrarily large elements of $T(x, y)$, $f(z) \leq 0$, so $T(x, y)$ cannot contain arbitrarily large positive real numbers. This is a contradiction to Claim 2. Hence, $f$ is non-decreasing. For any $0 < x < y$, we have $T(x, y) \cap [x, y]$ is dense in $[x, y]$ and $f(z) = mz + c$ for all $z \in T(x, y)$. Combined with the non-decreasing property of $f$, we have $f(z) = mz + c$ for all $z \in [x, y]$; i.e., $[x, y] \subseteq T(x, y)$. Now, it is easy to derive that $f$ is linear. Letting $M = f(2) - f(1)$ and $C = f(1) - M$, we have $f(z) = Mz + C$ for all $z \in [1, 2]$. Then, for $z \not \in [1, 2]$, consider an interval $[x, y]$ that contains both $z$ and $[1, 2]$. We have $f = z \mapsto mz + c$ on $[x, y]$, but $[1, 2] \subseteq [x, y]$ so $m = M$, $c = C$, and thus $f(z) = Mz + C$ as well. Hence, $f(z) = Mz + C$ for all $z \in \mathbb{R}^+$.

Consider all functions $f:\mathbb{R_+}\to \mathbb{R}$ which satisfy $$xf(x)+yf(y)=(x+y)f\left(\frac{x^2+y^2}{x+y}\right)$$.We claim that constant functions and linear functions are only solutions. $f(x)=1$ for all $x>0$ and $f(x)=x$ for all $x>0$ are 2 linearly independent solutions.Now note that solution set of such functions forms a vector space over field $\mathbb{R}$.Indeed $g$ , $h$ are 2 solutions then $ag+bh$ is also a solution for constant $a,b\in \mathbb{R}$.It is easy to check.If we can prove that this set has dimension 2 then clearly $ax+b$ will be the solution. So assume $f(1)=f(2)=1$.We will prove that $f$ is uniquely determined in each point. Claim:For all $x\in [1,2]$ we have $f(x)=1$ $\emph{proof:}$Simple competition gives $\frac{f(y)-f(x)}{y-x}=\frac{f(z)-f(y)}{y-x}$ where $z=\frac{x^2+y^2}{x+y}$ Now if there is a $a\in(1,2)$ with $f(a)<1$ then take $x_1$ such that $a=\frac{1+x_1^2}{1+x}$.Solution in $x_1$ exists since $a>1$.Again let $x_{n}$ be such that $x_{n-1}=\frac{x_n^2+1}{x_n+1}$.Since $\frac{(x_n-1)}{x_{n-1}-1}=\frac{(1+x_n)}{x_n}$ so $x_n$ will be large for large $n$. Now, $\frac{f(a)-f(1)}{a-1}= \frac{f(x_n)-f(1)}{x_n-1}$ gives $\frac{f(x_n)-f(1)}{x_n-1}$ is negative for large $n$.So $f(x_n)$ is negative.Contradiction. A similar argument shows that $f(x)$ can't be greater than 1 for $x\in [1,2]$.Here we will do the same thing forming $x_{n-1}=\frac{x^2+x_{n}}{x+x_n}$ and $f(2)=1$.$\square$ Claim: $f(x)$ is equal to 1 in $(a,b)$.Then $f(x)=1$ for all $x>a$. $\emph{proof:}$We will prove that there exists $\varepsilon>0$ such that $f(x)=1$ for all $x\in(a,b+\varepsilon)$. Indeed, for $r\to 0$ we have$\frac{(a+r)^2+(b+r)^2}{(a+r)+(b+r)}<b$.Fix a $r$ for which the inequality holds .Take $\varepsilon<r$.So for any $x\in [b,b+\varepsilon)$ we have $\frac{y^2+x^2}{y+x}\in (a,b)$ where $y=a+r$. Then $yf(y)+xf(x)=(x+y)f(\frac{y^2+x^2}{y+x})$ gives $xf(x)= x$.So $f(x)=1$. Repeat the same argument on $(a,b+\varepsilon)$.Observe that in this time as the length of the interval is increased to $b+\varepsilon-a$ from $b-a$ same $\varepsilon$ will work to make $f=1$ in $(a,b+2\varepsilon)$.$\square$ So $f(x)=1$ for all $x\ in [1, \infty)$ Now, take any $x<1$. For sufficiently large $y$ we have $\frac{y^2+x^2}{y+x}>1$. Apply $xf(x)+yf(y)=(x+y)f(\frac{y^2+x^2}{y+x})$ . So, $xf(x)+y=x+y$.Which gives $f(x)=1$. Hence $f\equiv 1$. Hence dimension of solution set of this kind of function is at most 2. So 1)$f(x)=ax+b$ for some constant $a,b\in R$. 2)$f(x)=c$ for all $x$. This indeed are solutions.$\blacksquare$ Now coming back to the main problem we are told to find all functions $f:\mathbb{R^+}\to {R^+}$ satisfying $$xf(x)+yf(y)=(x+y)f\left(\frac{x^2+y^2}{x+y}\right)$$So Indeed the solutions will be $\bullet$ Constant and $\bullet$ Linear.To avoid negative solutions we must have to take: $f(x)=ax+b$ for some constant $a,b\in \mathbb{R^+}$ $f(x)=c$ for some $c>0$. This are indeed solutions.$\square$ There is a problem in this post.Thanks to meagain for pointing out.

Informative problem! Solved with eisirrational and goodbear. The answer is $f(x) = ax+b$ with $a,b\geq 0$ and $(a,b) \neq (0,0)$. These are the only linear functions that work, so it remains to show $f$ linear. Define the following helper functions: $g(x,y) = \tfrac{x^2+y^2}{x+y}$, and $h(a,b)$ (for $a \leq b$) to be the unique $y$ for which $\tfrac{a^2+y^2}{a+y} = b$, which exists by IVT. We state the following lemmas in succession. Claim 1. Consider the line passing through $(a,f(a))$ and $(b,f(b))$. Then $(g(a,b), f(g(a,b)))$ and $(h(a,b), f(h(a,b)))$ also lie on this line. Proof. This is clear from the assertion. A way to think about this: given $f(a), f(b)$, the value of $f(g(a,b))$ and $f(h(a,b))$ are uniquely determined. Then since $f$ linear is a solution, these two points must lie on this linear function. $\square$ Claim 2. The function $f$ is nondecreasing. Proof. Suppose $a < b$ and $f(a) > f(b)$. Consider the iterative sequence \[h(a,b), \quad h(a,h(a,b)), \quad h(a,h(a,h(a,b))), \dots,\]which is strictly increasing. It either diverges or has some supremum $L$; for this supremum, $h(a,L) = L$. But this is not true for any $L > a$, so this sequence diverges. The line through $(a,f(a))$ and $(b,f(b))$ dips below $0$ for all $x > N$ for some $N$. Then just select a number from the above sequence that is greater than $N$, which will have negative image and produce a contradiction. $\square$ Claim 3. The function $f$ is linear. Proof. We show $f$ is linear on any interval $[a,b]$, which is enough. Let $S$ be the set of numbers such that $a,b \in S$, and if $x,y \in S$ then $g(x,y) \in S$. Since $x < g(x,y) < y$, this set is dense. Moreover, $f$ is linear in $S$, so because $f$ is nondecreasing it is linear in $[a,b]$ as desired. $\square$

Time to fill in all the details: I claim the only functions that work are $f(x)\equiv dx+c$ where $d,c\in \mathbb{R}_{\ge 0}$. Clearly they work, and now I claim they're the only functions. Let $f(1)=c$ and $f(2)=c+d$. I claim $f(x)\equiv dx-(d-c)$. Call $x$ good if $f(x)=dx-(d-c)$. First I show the set of good reals is dense. Initially, $x=1,y=2$. We first "fill" the interval $(x,y)$, meaning for any good number $p$ in the interval $(x,y)$ and real number $\epsilon>0$, there exist another $q$ good such that $0<q-p<\epsilon$ and $r$ good such that $0<p-r<\epsilon$. Note if $x,y$ are good then $\frac{x^2+y^2}{x+y}$ is good. Similarly, if $x,\frac{x^2+y^2}{x+y}$ are good then $y$ is good. We fill in a divide-and-conquer method. In this process, we always assume $x<y$. Note $x<\frac{x^2+y^2}{x+y}<y$, so $\frac{x^2+y^2}{x+y}$ is good and we only need to "fill" the intervals $(x,\frac{x^2+y^2}{x+y})$ and $(\frac{x^2+y^2}{x+y},y)$. We can repeat this process infinitely many times to meet our definition of density. Then if $x$ is good, $y$ is good, $0<x<y$, then we find $z$ such that $\frac{x^2+z^2}{x+z}=y$. It exists because it is a root of $z^2-yz-(yx-x^2)=0$ and is greater than $y$. We can get $f(z)=dz-(d-c)$, and we now fill the interval $(y,z)$. Solving the quadratic we can see $z\le y\frac{1+\sqrt{2}}{2}\le y\frac{13}{10}$ We repeat this operation in this manner: find the root of $\frac{z}{2}+\varepsilon$ that is good, which works because $z\le y\frac{13}{10}$ and good numbers in the interval $(1,y)$ is dense. Now we have $yx-x^2>\frac{y^2}{4}-\delta$ for some $\delta$ arbitarily small (just control $\varepsilon$), so $z=\frac{y+\sqrt{2y^2-4\delta}}{2}>\frac{12}{10}y$. Therefore I can fill up to infinity. Now I fill the interval $(0,1)$. We pick $f(z)=dz-(d-c)$ and $\frac{k^2+z^2}{k+z}=1$. This gives $k=\frac{1+\sqrt{1-4(z^2-z)}}{2}$ We pick $z=\frac{1+\sqrt{2}}+\epsilon$ such that $z^2-z>\frac 14-\delta$. Hence $k<\frac{1+\sqrt{4\delta}}{2}$. I can fill the interval $(k,1)$ in the previous manner, and continue doing this until $k<\epsilon$. Hence good numbers are dense in $(0,+\infty)$ If $d<0$, we can pick $x$ arbitarily large such that $x>\frac{c}{-d}$ and $f(x)=dx-c<0$, so $d\ge 0$. If $d-c<0$, we pick $x$ arbitarily small such that $x<\frac{-c}{d}$ and $f(x)=dx-(d-c)<0$. Now we let $k=c-d$ Now suppose there exist $x$ such that $f(x)>dx+k$. We pick $\varepsilon\rightarrow 0$ small, such that $\frac{\varepsilon^2+x^2}{\varepsilon+x}$ is good and $(x+\varepsilon)(d\frac{\varepsilon^2+x^2}{\varepsilon+x}+k)<xf(x)$, we get $f(\varepsilon)<0$, false. If $f(x)<dx+k$, we find $y$ such that $f(y)$ is good and the $z$ such that $\frac{x^2+z^2}{x+z}$ has $f(z)>dz+k$.

Th3Numb3rThr33 wrote: Let $S$ be the set of numbers such that $a,b \in S$, and if $x,y \in S$ then $g(x,y) \in S$. Since $x < g(x,y) < y$, this set is dense. I don’t think this logically follows; this claim requires some more accurate bounding.

Somewhat strange solution Some housekeeping first: the answer is all linear function $f(x)=\alpha x + \beta$, where $\alpha,\beta\geq 0$ but not both are equal to $0$. They obviously satisfy the functional equation, therefore we restrict our attention into proving that these are all solutions. Let $P(x,y)$ denote the functional equation. Our first step is to do Vieta Jumping on the functional equation. The main result from this is the following. Claim: $af(a) - bf(b) = (a-b)f(a+b)$ for any $a,b\in\mathbb{R}^+$. Proof: Fix $t,k\in\mathbb{R}^+$ where $t<k$. Let $g(x)=\tfrac{x^2+t^2}{x+t}$. Since $\lim_{x\to\infty}g(x)=\infty$ and $g(t)=t$, there exists $a$ such that $g(a) = \tfrac{t^2+a^2}{t+a}=k$. Moreover, by Vieta's Jumping, $g(k-a)=k$. Thus, by considering $$\begin{array}{rcrcl} P(t,a) & \implies & tf(t) + af(a)&=&(a+t)f(k)\qquad\text{ and} \\[4pt] P(t,k-a) & \implies & tf(t) + (k-a)f(k-a)&=&(k-a+t)f(k), \end{array}$$we get that $$af(a) - (k-a)f(k-a) = (2a-k)f(k)$$for any $a,k$ such that $k>a$. Thus, by setting $b=k-a$, we get the desired claim. $\blacksquare$ Now let's do some geometry. Let $P_x = (x,f(x))$. Then the claim above implies that $P_a, P_b, P_{a+b}$ are colinear whenever $a,b\in\mathbb{R}^+$. Fix any pairwise distinct $a,b,c\in\mathbb{R}^+$. Set $A=P_a$, $B=P_b$, $C=P_c$, $D=P_{b+c}$, $E=P_{c+a}$, and $F=P_{a+b}$. By the claim, $D\in BC$, $E\in CA$, $F\in AB$, and $AD,BE,CF$ are concurrent at $P_{a+b+c}$. However, we claim that Claim: $D,E,F$ are colinear. Proof: If $A,B,C$ are colinear, then the result is trivial. Otherwise, we use Menelaus theorem. Observe that $$\frac{\overline{BD}}{\overline{DC}} = \frac{b - (b+c)}{(b+c)-c} = -\frac bc$$hence multiplying the result cyclically gives the result. $\blacksquare$ By Ceva's theorem, both $D,E,F$ being colinear and $AD,BE,CF$ being concurrent can only happen when $A,B,C$ are colinear. This means that $(a,f(a))$, $(b,f(b))$, $(c,f(c))$ are colinear for any $a,b,c\in\mathbb{R}^+$. This concludes, say by varying $c$, that $f$ must be linear.

MarkBcc168 wrote: Since $\lim_{x\to\infty}f(x)=\infty$ and $g(t)=t$, . I think that you want to say $\lim_{x\to\infty}g(x)=\infty$