Does this look right?
$\log \prod\limits_{i=1}^{m}{\left( \begin{align} & {{n}_{i}} \\ & {{y}_{i}} \\ \end{align} \right)\theta _{i}^{{{y}_{i}}}{{(1-{{\theta }_{i}})}^{{{n}_{i}}-{{y}_{i}}}}}$
$=\log \prod\limits_{i=n}^{n}{{{\theta }_{i}}^{{{y}_{i}}}}+\log \prod\limits_{i}^{n}{{{(1-{{\theta }_{i}})}^{{{n}_{i}}-{{y}_{i}}}}}$ $\text{Note: I ignored the therm } \left( \begin{align} & {{n}_{i}} \\ & {{y}_{i}} \\ \end{align} \right)$ $=\sum\limits_{i=1}^{m}{{{y}_{i}}\log {{\theta }_{i}}}+\sum\limits_{i=1}^{m}{({{n}_{i}}-{{y}_{i}})\log (1-{{\theta }_{i}})}$
Calculating first derivative:
$\frac{\partial }{\partial \theta }=\frac{\sum{{{y}_{i}}}}{{{\theta }_{i}}}-\frac{\sum{({{n}_{i}}-{{y}_{i}})}}{1-{{\theta }_{i}}}=0$ $\frac{\sum{{{y}_{i}}}}{{{\theta }_{i}}}=\frac{\sum{({{n}_{i}}-{{y}_{i}})}}{1-{{\theta }_{i}}}$ $\sum{{{y}_{i}}-{{\theta }_{i}}\sum{{{y}_{i}}={{\theta }_{i}}\sum{({{n}_{i}}-{{y}_{i}})}}}$ $\sum{{{y}_{i}}={{\theta }_{i}}\left( \sum{({{n}_{i}}-{{y}_{i}})+\sum{{{y}_{i}}}} \right)}$ $\sum{{{y}_{i}}={{\theta }_{i}}(\sum{{{n}_{i}}})}$ ${{\hat{\theta }}_{i}}=\frac{\sum{{{y}_{i}}}}{\sum{{{n}_{i}}}}$