$\newcommand\th\theta\newcommand\si\sigma\newcommand\p\partial\newcommand\ol\overline$There is no reason to get confused here. Indeed, that "the first term [whatever it means -- I.P.] of the derivative does not depend on $(z_j-\mu_j(\theta))$" does not at all prevent the derivative from taking the zero value.

If e.g. $\mu(\th)=\th$ and $\si_i(\th)=\th$ for all real $\th>0$ and all $i=1,\dots,m$, then $$\frac{\p L}{\p\th}=-\frac n{\th^3}\,(-\ol{z^2}+\th\ol z+\th^2),$$ where $\ol z:=\frac1m\,\sum_{i=1}^m z_i$ and $\ol{z^2}:=\frac1m\,\sum_{i=1}^m z_i^2$, so that $\frac{\p L}{\p\th}=0$ at a real $\th>0$ if and only if $$\th=\hat\th:=\hat\th_m:=\frac{-\ol z+\sqrt{\ol z^2+4\ol{z^2}}}2.$$

Moreover, here $\hat\th$ is the maximum likelihood estimator of $\th$. Using (say) the law of large numbers, one can easily check that $\hat\th_m$ is consistent: $\hat\th_m\to\th$ in probability as $m\to\infty$ (assuming that $\th$ is the true value of the parameter).

$\newcommand\th\theta\newcommand\si\sigma\newcommand\p\partial\newcommand\ol\overline$There is no reason to get confused here. Indeed, that "the first term of the derivative does not depend on $(z_j-\mu_j(\theta))$" does not at all prevent the derivative from taking the zero value.

If e.g. $\mu(\th)=\th$ and $\si_i(\th)=\th$ for all real $\th>0$ and all $i=1,\dots,m$, then $$\frac{\p L}{\p\th}=-\frac n{\th^3}\,(-\ol{z^2}+\th\ol z+\th^2),$$ where $\ol z:=\frac1m\,\sum_{i=1}^m z_i$ and $\ol{z^2}:=\frac1m\,\sum_{i=1}^m z_i^2$, so that $\frac{\p L}{\p\th}=0$ at a real $\th>0$ if and only if $$\th=\hat\th:=\hat\th_m:=\frac{-\ol z+\sqrt{\ol z^2+4\ol{z^2}}}2.$$

Moreover, here $\hat\th$ is the maximum likelihood estimator of $\th$. Using (say) the law of large numbers, one can easily check that $\hat\th_m$ is consistent: $\hat\th_m\to\th$ in probability as $m\to\infty$ (assuming that $\th$ is the true value of the parameter).