Let $P(x,y,z): f(x)f(y)f(z)=9f(z+xyf(z)) P(x,\frac{1}{x},z): f(x)f(\frac{1}{x})f(z)=9f(z+f(z))$ which is true $\forall x,z\in\mathbb{R^{+}} \therefore f(x)f(\frac{1}{x})=c$ and $9f(x+f(x))=cf(x)$ for some constant $c$. Now $P(x,\frac{1}{y},y): cf(x)=9f(y+\frac{x}{y}f(y))$ Now take $x$ close to $0$. This means that $9f(y)=cf(x) \forall y\in\mathbb{R^{+}}$ which means that $f$ is constant. Let $f(x)=d$. We have $d^3=9d$ which gives $d=-3,0,3$ but only $3$ is accepted therefore: $f(x)=3 \forall x\in\mathbb{R^{+}}$

How can you take x close to 0 maybe the function is 10^infinite so you will have xf(z) will be greater then 0

Here is a solution which seems to work. Let $P (x,y,z) $ correspond to the given assertion. Comparing $P (x,y,z)$ and $P (xy,1,z) $ we conclude that $f (x)f (y)=f (xy)f (1) $. Thus we see that $f (x)f (y)f (z)=f (xy)f (z)f (1)=f (xyz)f (1)^2$. Also $f (z+xyf (z))f (1)=f (xyz)f \left(\frac {1}{xy}+\frac {f (z)}{z}\right) $. Putting these values in $P (x,y,z)$ gives $f (xyz)f (1)^3=9f (xyz)f \left(\frac {1}{xy}+\frac {f (z)}{z}\right) $ $\implies f (1)^3=9f \left(\frac {1}{xy}+\frac {f (z)}{z}\right) $. Fixing $z $ and letting $x,y $ range over $\mathbb {R}^+$, we see that $\forall t\in\mathbb {R}^+,f\equiv\frac {f (1)^3}{9} $ over the interval $\left (\frac {f (t)}{t},\infty\right) $. In particular, $f\equiv \frac {f (1)^3}{9} $ over the interval $\left (f (1),\infty\right) $ using $t=1$. Taking $x,y,z $ in that interval $P (x,y,z)\implies f (1)=3\implies f\equiv 3$ over the interval $\left ( f (1),\infty\right) $. Now in $P (x,y,z) $, take $y,z>f (1) $, then $f (y)=f (z)=3$. Also $z+xyf (z)>z>f (1)\implies f (z+xyf (z))=3$, so now $P (x,y,z)\implies f (x)=3\forall x\in\mathbb {R}^+$. Thus the only solution is $f\equiv 3$.

I have an extremely simple solution: Let $P (x,y,z) $ correspond to the given assertion. Then, $P(0,0,z)$ corresponds to $f(0)^2f(z)=9f(z)$, and therefore, $f(0)=3.$ Now, notice that $P(x,0,0)$ corresponds to $f(x)f(0)^2=9f(0)$, meaning that $9f(x)=9f(0)$, and $f(x)$ is a constant function, and $f(x)=3$ for all real $x.$

Plops wrote: Then, $P(0,0,z)$ corresponds to $f(0)^2f(z)=9f(z)$ You are not allowed to plug in $0$ since $0$ is not in the domain of $f$ and $x,y,z$ are positive reals.

Smh. I knew it was too simple of a solution.

Pick following such that x>z. $(x, \frac{x-z}{xf(z)}, z)\implies f(\frac{x-z}{xf(z)})f(z)=9$. So, f(k) is constant for k in $(0,\frac{1}{f(z)})$. Now, for z and z' we pick a k such that 0<k< $\frac{1}{f(z)}$ and 0<k<$\frac{1}{f(z')}$. Then, f(z)=$\frac{9}{f(k)}=f(z')$. So, f(z) is a constant. So we get $c^2=9$. So f(x)=3

Rg230403 wrote: Pick following such that x>z. $(x, \frac{x-z}{xf(z)}, z)\implies f(\frac{x-z}{xf(z)})f(z)=9$. So, f(k) is constant for k in $(0,\frac{1}{f(z)}$. Now, for z and z' we pick a k such that 0<k< $\frac{1}{f(z)}$ and 0<k<$\frac{1}{f(z')}$. Then, f(z)=$\frac{9}{f(k)}=f(z')$. So, f(z) is a constant. So we get $c^2=9$. So f(x)=3 But that just implies it is constant over some interval, not the whole of $\mathbb{R}^+$ @below: OK, thanks, got it

Supercali wrote: Rg230403 wrote: Pick following such that x>z. $(x, \frac{x-z}{xf(z)}, z)\implies f(\frac{x-z}{xf(z)})f(z)=9$. So, f(k) is constant for k in $(0,\frac{1}{f(z)}$. Now, for z and z' we pick a k such that 0<k< $\frac{1}{f(z)}$ and 0<k<$\frac{1}{f(z')}$. Then, f(z)=$\frac{9}{f(k)}=f(z')$. So, f(z) is a constant. So we get $c^2=9$. So f(x)=3 But that just implies it is constant over some interval, not the whole of $\mathbb{R}^+$ No because z and z' are arbitrary and we have f(z)=f(z')

let $P(x,y,z)$ be the assertion of $f(x)f(y)f(z)=9f(z+xyf(z))$ $P(xy,1,z)$ comparing with $P(x,y,z)$ we get $f(xy)=\frac{f(x)f(y)}{f(1)}$ $(*)$ $P(x,y,1):f(x)f(y)f(1)=9f(xyf(1)+1)$ so $f(xy)f(1)^2=9f(xyf(1)+1)$ ,$f(x)f(1)^2=9f(xf(1)+1)$ plugging $x$ with $\frac{x}{f(1)}$ yields $f(\frac{x}{f(1)})f(1)^2=9f(x+1)$ from $(*)$, we have $f(\frac{x}{f(1)})=f(x \cdot \frac{1}{f(1)})=\frac{f(x)f(\frac{1}{f(1)}}{f(1)}$ Let $c=\frac{f(1)\cdot f(\frac{1}{f(1)})}{9}$ then $f(x+1)=cf(x)$ Now plugging $y$ with $y+1$ in $(*)$ gives us $f(xy+x)=\frac{f(x)f(y+1)}{f(1)}=\frac{cf(x)f(y)}{f(1)}$ and comparing with $(*)$ we have $f(xy+x)=cf(xy)$ i.e. $f(x+1) = cf(1)$ so $f$ is constant and so $f(x)=3$ for every positive real numbers $x$

x=1:f(z) so:) f(1:f(z))=f(1:f(y)) easily and f must be constant.