Note: $k=N$ throughout the answer.
It is well-known that the standard chi-squared distribution with $k$ degrees of freedom has probability density function $$ f_{\sum_{i=1}^{N}Z_i^2}(z)= \frac{z^{\frac{k}{2}-1} e^{-\frac{z}{2}}}{ 2^{\frac{k}{2}}\Gamma(\frac{k}{2})} \mathbb{I}_{z>0}.$$
We may write $Y=\sum_{i=1}^{N}(\sigma Z_i)^2=\sigma^2 \sum_{i=1}^{N}Z_i^2$, where the $Z_i$'s are independent and identically distributed standard normal variables. ($Z_i=\frac{X_i}{\sigma}$ for all i.)
By the change of variables formula applied to the monotonic function $g(y)=\sigma^2 y$, we have, $$f_{Y}(y)= \frac{1}{\sigma^2}f_{\sum_{i=1}^{N}Z_i}(\frac{y}{\sigma^2})= \frac{1}{\sigma^2} \frac{ (\frac{y}{\sigma^2})^{\frac{k}{2}-1} e^{-\frac{\frac{y}{\sigma^2}}{2}}}{ 2^{\frac{k}{2}}\Gamma(\frac{k}{2})} \mathbb{I}_{y>0}.$$
See https://en.wikipedia.org/wiki/Probability_density_function for a reminder on the change of variables formula.