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So far what I can think of on this question is: Rewrite $u(f(x,y))=u(x) \Psi(\frac{w(x,y)}{u(x)})$ as: $\frac {u(f(x,y))}{u(x)}=\Psi(\frac {w(x,y)}{u(x)})$ Since $\Psi$ can be arbitrary, we can further rewrite it as: $\Phi(\frac {u(f(x,y))}{u(x)})=\Gamma({\frac {w(x,y)}{u(x)})$ where $\Psi=\Phi^{-1}(\Gamma)$ Above equation shows that any $w(x,y),u(x)$ which makes $(\frac{w(x,y)}{u(x)}$ span into $(\Phi(frac{f(x,y)}{u(x)})$ is a solution. For example, given $f(x,y)=ax+by$, select $u=x$, then: $\Phi(frac{f(x,y)}{u(x)})= \Phi(a+b\frac{y}{x})$ which can be spanned by $\frac {y}{x}$, so $w=y$ is a solution. Question: how to find the appropriate $w(x,y),u(x)$ to span into $\Phi(\frac {u(f(x,y))}{u(x)})$