Are maximum likelihood estimator robust estimators? It seems to me that since 
$$
\widehat{\vec{\theta}} = \mathrm{argmin}_{\vec{\theta}} \sum_{i=1}^{n} - \log(f(x_i; \vec{\theta})) = \sum_{i=1}^{n} - \log\left( \frac{\partial F}{\partial x}(x_i; \vec{\theta} )\right) = \widehat{\vec{\theta}}(F),
$$
that this means that maximum likelihood estimators are robust estimators. Is this correct?
 A: By the definition of robust estimators, this is true. That is, M-estimators are a type of robust statistics, and MLE's are a special case of M-estimators. 
However, it's definitely not the case that MLE's in general have good robustness properties. 
A: One way to look at it is that there is no strict distinction between robust and non-robust estimators, but rather they can be compared according to their robustness properties (as already pointed in the answer by @CliffAB.) Thus, M-estimators can be viewed as explicitly designed to resemble the maximum likelihood estimators (see, e.g., their introduction in Robust Statistics: Theory and Methods by Maronna, Martin and Yohai.). That is, M-estimators are defined as minima of a function of the type shown in the OP (or as zeros of a derivative of such a function), which however does not necessarily have an interpretation of likelihood (e.g., the associated probability density $f(x; \vec{\theta})$ might not be normalizable.)
Thus, both sample mean and sample median can be viewed as M-estimators for location, but the former with breakdown point of 0 and the latter with breakdown point of 50%.
This suggests the other way to look at: by defining the non-robust estimators as those with extremely poor measures of robustness. Thus, we could define as non-robust any estimator with breakdown point of 0, meaning that even a single outlying observation may produce arbitrarily big deviation from the true value of the parameter being estimated.
