The answer is $f(x) = c ^ { \frac{1}{lnx} } $, where $ c>1 $ is an arbitrary constant. Let's go ahead and prove it. Consider a function $g$ such that $f(x) = g(x) ^ { \frac{1}{lnx} }$. Obviously, such a $g(x)$ exists for any $x>1$ and it's unique. Let's take $y=x$ in the initial condition and write that in terms of $g$: $g(x^{u+v}) ^ { \frac {1} {(u+v)lnx} } \leq { g(x)^{\frac{1}{4lnx} (\frac{1} {u} +\frac{1}{v} ) } } $, which can be rewritten as: $g(x^{u+v}) \leq {g(x)^ {\frac{1}{4} (u+v) (\frac{1}{u}+\frac{1}{v}) } } $. Now take $u=v$, for arbitrary $v$. The last inequality becomes: $g(x^{2v}) \leq {g(x)} $. As this holds for any $v>0$, we have that $g$ is a constant function, and $g(x)=c$ holds for any $x>1$, after which the result follows. What's left is to check that $f$ is actually a solution. The proof is straightforward once you use the facts that $f(x)>1$ and $(u+v)(\frac{1}{4u} + \frac{1}{4v}) \geq {1}$, for any $u,v>0$.