Suppose we have a parameter $\theta \in R^{n}$ that defines some noisy observation $z=\mu(\theta)+\eta, z\in R^{m}$ where the noise follows a Gaussian distribution whose covariance is a function of the parameter, i.e. $\eta\sim N(0,\Sigma(\theta))$. Then the observation random variable is $z \sim N(\mu(\theta),\Sigma(\theta))$. Assume observations of $z$ are iid and $\Sigma=diag(\sigma_1^{2}(\theta), \sigma_2^{2}(\theta)...\sigma_m^{2}(\theta))$, where $\sigma_i^{2}\ne\sigma_j^{2}$.

Then, the log-likelihood function would be:

\begin{align} L_z\left(z;\theta\right)&=log\left(\prod^m_{j=1}\frac{1}{\sqrt{2 \pi \sigma_j^2\left(\theta\right)}} e^{-\frac{\left(z_j-\mu_j(\theta)\right)^2}{2 \sigma_j^2\left(\theta\right)}}\right)\\ &=\sum_{j=1}^{m} -\frac{1}{2}log(2\pi)-\frac{1}{2} log(\sigma_{j}^{2}(\theta))-\frac{(z_j-\mu_j(\theta))^{2}}{2\sigma_j^{2}(\theta)}\\ \end{align}

The first derivative of the log-likelihood function w.r.t the parameters would be:

\begin{aligned} \frac{\partial L}{\partial \theta}= \sum_{j=1}^{m}-\frac{1}{2}\left(\sigma_j^2(\theta)\right)^{-1} \frac{\partial \sigma_j^2}{\partial \theta}-(z_j-\mu_j(\theta)) \frac{\partial \mu_j}{\partial \theta}\left(\sigma_j^2(\theta)\right)^{-1}-\frac{1}{2}\left(\sigma_j^2(\theta)\right)^{-2} (z_j-\mu_j(\theta))^2 \frac{\partial \sigma_j^2(\theta)}{\partial \theta} \end{aligned}

What I'm getting confused about is: how would the log-likelihood function derivative be zero at the parameter value, when the first term of the derivative does not depend on $(z_j-\mu_j(\theta))$?