This isn't necessarily true. Let $\phi(x) = \cos(x)$ and $\psi(x) = \cos(2x)$. Let $f(x)$ be a complex-valued function such that $|f(x)|$ has its absolute maximum at some $\xi$ for which the (complex-valued) derivative $f'(x)$ is nonzero at $\xi$. Then one can integrate by parts to get
$$\int_0^{\pi} \cos(x)f(x)^n dx = n\int_0^{\pi}\sin(x)f'(x)f(x)^{n-1} dx$$
$$\int_0^{\pi} \cos(2x)f(x)^n dx = {n \over 2}\int_0^{\pi}\sin(2x)f'(x) f(x)^{n-1}dx$$
If your statement were true, then by looking at the ratio of the left-hand sides and taking limits as $n$ goes to infinity one should get ${\cos(\xi) \over \cos(2\xi)}$, while looking at the ratio of the right-hand sides and taking limits as $n$ goes to infinity one should get ${2 \sin(\xi) \over \sin(2\xi)}$, which is generally not the same.