For maximum likelihood estimation, do we theoretically need to check that the critical point is a maximum (rather than a minimum or saddle point) or is this automatic?
I believe that it is automatic and it has to do with the fisher information matrix, which comes from the second partial derivatives matrix, which we know is the the inverse of a variance matrix, which we know is semi-positive definite.
Counterxample: Let $X$ be distributed as a mixture of two variables $Y,Z$ with some known densities and mixing paramenter $\theta \in [0,1]$. Then $p_X(x_i) = \theta \, p_Y(x_i)+(1-\theta)\, p_Z(x_i) = \theta \,g(x_i)+h(x_i)$ and
$$L(\theta)=\prod_{i=1}^n (\theta \,g(x_i)+h(x_i))$$
This is a polynomial of degree $n$ in $\theta$, so it can have several minima and maxima inside the domain - and it's easy to construct an example.
So, in general, you need to check that the critical point is indeed a maximum.
Furthermore (and practically more relevant), the global maximum is not always a critical point, so computing the derivative of the likelihood and equation to zero is not always the right way. You need to check that it's indeed differentiable in all its domain, and you need to check the boundaries. The classical example is a uniform distribution in $[0,\theta]$.
(Notice that the above remarks are not specific to a likelihood function, they apply when looking at the global maximum of any function.)
One more relevant thing: the likelihood often has several local maxima, a fact to take into account when using numerical iterative methods for finding the MLE. Now, if there are several maxima, and if the likelihood is differentiable, then one should expect also other critical points that are not maxima.
The properties of the Fisher information matrix are not relevant here, because they require some regularity conditions and, more crucially, because they deal with an expectation of the derivative of the likelihood. To know that the expectation of something is positive does not tell that that something is positive.