Prove Neg. Log Likelihood for Gaussian distribution is convex in mean and variance.

So you get $$l(\mu,\alpha) =\frac{n}{2}\ln 2 \pi - \frac{n}{2} \ln \alpha+ \sum \frac{(x_i- \mu)^2\alpha}{2}$$ Convex in $\mu$

The second derivative w.r.t $\mu$ is $$\frac{\partial^2}{\partial \mu^2}l = n \alpha > 0$$ So we get convexity in $\mu$.

Convex in $\alpha$

The second derivative w.r.t $\alpha^2$ is $$\frac{\partial^2}{\partial \alpha^2}l = \frac{1}{\alpha^2} > 0$$ So we get convexity in $\alpha$.

What I think you meant is that you would want to prove that $l(\pmb{z})$ is convex in $\pmb{z}$, where $\pmb{z} = [\mu, \alpha]$ (jointly). Well, it is not convex in $\pmb{z}$ because the Hessian you wrote has negative values for values of $x_i,\mu,\alpha$: Choose a small $\frac{2}{\alpha}$ and a large $4(x_i - \mu)^2$, this leaves us with a negative determinant. Boyd does not tell you that $l(\mu,\alpha)$ is convex in $\mu,\alpha$. The statement convex in mean and variance means that it is convex in mean and it is convex in variance.

The link you shared here is something completely different. They want to show that the optimal values are concave (at least this is what they state).