Let $x>z>0$ be arbitrary. Set $y=\frac{x+z}{2}$, then $x>y>z$ applies and the equation simplifies to

$$f(x)+f(z)=\frac{x^{2}+z^{2}}{2}, \quad \forall x>z>0$$

If we set $z=1$ and $x>1$ on the one hand, and $x=1$ and $00, x \neq 1
If, for example, we now set $x=2$ and $y=3$ in (2) and use what has just been proven, then finally $f(1)=\frac{1}{2}$ follows and therefore $f(x)=\frac{x^{2}}{2}$ is the only possible solution function. This obviously fulfills the condition of the task even for all $x, y, z>0$. ## 2nd solution If you first set $x=2 t, y=\frac{3}{2} t, z=t$, then $x=3 t, y=2 t, z=t$, and finally $x=3 t, y=\frac{5}{2} t, z=2 t$ for any $t>0$, you get the three equations
$$\begin{aligned} f(2 t)+f(t) & =\frac{5}{2} t^{2} \\ f(3 t)+f(t) & =5 t^{2} \\ f(3 t)+f(2 t) & =\frac{13}{2} t^{2} \end{aligned}$$
If you add the first two and subtract the third, then $2 f(t)=t^{2}$ follows for all $t>0$, the only solution function is therefore $f(t)=\frac{t^{2}}{2}$. ## 3rd solution Set $x=3 t, y=2 t, z=t$ for $t>0$, then follows
f(3 t)+f(t)=5 t^{2}
Set $x=9 t, y=5 t, z=t$ for $t>0$, then you get analogously
f(9 t)+f(t)=41 t^{2}
Finally, substitute $x=9 t, y=3 t, z=t$ into the original equation and multiply by 2 , then the following follows
$$\begin{aligned} 2 f(7 t) & =(f(9 t)+f(3 t))+(f(9 t)+f(t))+(f(3 t)+f(t))-42 t^{2} \\ & =45 t^{2}+41 t^{2}+5 t^{2}-42 t^{2}=49 t^{2} \end{aligned}$$
$$$$

If we replace $7 t$ with $t$, we finally obtain the only possible solution $f(t)=\frac{t^{2}}{2}$ for $t>0$.