By setting $P(x)=e^x, g(x)=ln(x), h(x,y)=ln(y)$, we have: $p(x,y)=xy=e^{ln(xy)}=e^{ln(x)+ln(y)}=P(g(x)+h(x,y))$ Where: $\frac {\partial lnx}{\partial x}=\frac {1}{x}>0$, $\frac {\partial lny}{\partial y}=\frac {1}{y}>0$ for $x>0, y>0$.

Yes. This yields one solution. However any more functions other this? I gave one example in belowing post, hope to have your advices?

Hi Willie, Thanks for the advise. I intended to do so but found the words exceeded the limit, so I changed to use the "Add Another Answer". But I will keep your advice in mind.