The closest line of work to your problem is
Multi-View Clustering. Each data set DS is considered as a view, and views share a central entity (e.g. an individual).
Multi-View Clustering is a survey on the topic. Also, here are two git projects from two papers (git 1, git 2) on the topic (they are implemented in Matlab and unfortunately are not production-ready).
Since most mainstream clustering methods (e.g. K-means, DBSCAN) accept a custom distance matrix $d(x, y)$ instead of explicit data vectors $x$ and $y$, partiality of features can be handled outside the method.
For example, suppose $x$ and $y$ are both present in views $1$ and $3$ but not view $2$. You may first build their vector by fusing their features from views $1$ and $3$, then computing their distance $d(x, y)$. Do the same for all pairs of individuals based on their common features, and finally feed the distance matrix $d(X, Y)$ to a clustering algorithm. This way, based on your domain knowledge, you can manually assign different weights to features of different views too. For example, you can define the distance as
$$d(x, y) := \left \| x[f_1:f_3] - y[f_1:f_3] \right \| + 2\left \| x[f_4:f_5] - y[f_4:f_5] \right \|$$
which implies that the difference in view $3$ with features $f_4$ and $f_5$ is twice as important as the difference in view $1$ with features $f_1$ to $f_3$.
Comparing two partitions
If views are clustered separately, two partitions $p_1$ and $p_2$ respectively
over views $1$ and $2$ can be compared using metric
Normalized Variation of Information (NVI). If partitions are the same, it outputs $0$, and if they are statistically independent, it outputs $1$. A similar measure (that is not a metric) is
Normalized Mutual Information (NMI).
If all individuals are not present in both partitions, either (1) remove non-shared individuals from both partitions, or (2) individuals in $p_1$ that are not present in $p_2$ can be assigned to an imaginary cluster $misc_2$ in $p_2$, the same for those in $p_2$ that are not in $p_1$ using $misc_1$ in $p_1$.