Asymptotic complexity of function with two Input variables
No, it's not safe to say either.  The intuition: When $n$ is really large, $f(m,n)$ grows faster than $O(m \log n)$.  When $m$ is really large, $f(m,n)$ grows faster than $O(n^2)$.

Instead, what you can say is that $f(m,n)$ is $O(m \log n + n^2 + n \log \log m)$.  Why?

$$\begin{align*}
f(m,n) &amp;= \log {n^m} + 100n \log \log {m^5} + 150m + 4n^2 + 1000\\
  &amp;=m \log n + 100n\log \log m + 100n\log 5 + 150m + 4n^2 + 1000\\
  &amp;=O(m \log n) + O(n \log \log m) + O(n) + O(m) + O(n^2) + O(1)\\
  &amp;=O(m \log n) + O(n \log \log m) + O(n^2) + O(m \log n) + O(n^2) + O(1)\\
  &amp;=O(m \log n + n \log \log m + n^2).
\end{align*}$$

If $n \log \log m = O(m \log n + n^2)$ (I'm not sure if this is true), then you could simplify this to $O(m \log n + n^2)$.