Problems like this, where you want to differentiate the product of a bunch of functions that depend on your variable of interest, can be dealt with by logarithmic differentiation. Let $\Phi$ and $\phi$ denote the CDF and PDF of the standard normal distribution (respectively). Since the normal random variables in your question have the same variance you get:
$$\prod_{i=1}^n F_i(y) = \prod_{i=1}^n \Phi \Big( \frac{y-\mu_i}{\sigma} \Big) = \exp \Bigg( \sum_{i=1}^n \ln \Phi \Big( \frac{y-\mu_i}{\sigma} \Big) \Bigg).$$
Differentiating with respect to $y$ and applying the chain rule gives:
$$\begin{equation} \begin{aligned}
f_Y(y) = \frac{d F_Y}{dy}(y)
&= \Bigg( \frac{1}{\sigma} \sum_{i=1}^n \frac{\phi ( (y-\mu_i)/\sigma ) }{\Phi ( (y-\mu_i)/\sigma )} \Bigg) \exp \Bigg( \sum_{i=1}^n \ln \Phi \Big( \frac{y-\mu_i}{\sigma} \Big) \Bigg) \\[6pt]
&= \Bigg( \frac{1}{\sigma} \sum_{i=1}^n \frac{\phi ( (y-\mu_i)/\sigma ) }{\Phi ( (y-\mu_i)/\sigma )} \Bigg) \Bigg( \prod_{i=1}^n \Phi \Big( \frac{y-\mu_i}{\sigma} \Big) \Bigg). \\[6pt]
\end{aligned} \end{equation}$$
In the special case where $\mu = \mu_1 = \cdots = \mu_n$ this reduces to the well-known formula:
$$f_Y(y) = \frac{n}{\sigma} \cdot \phi \Big( \frac{y-\mu}{\sigma} \Big) \cdot \Phi \Big( \frac{y-\mu}{\sigma} \Big)^{n-1}.$$