If you measure one "share" of an entangled pair, will the resulting pair be a product state? If you do a partial measurement on one "share" of en entangled pair, will the resulting pair no longer be entangled, i.e will be a product state?
 A: It depends on what you mean precisely with "measuring", but generally speaking, no.
If you are talking about a projective measurement, and you are asking about the state of the rest of the system conditionally to the measurement result, then sure the residual state is separated from the measured one.
More precisely, this interpretation of the question amounts to asking the state obtained after performing a partial projection. If the initial state is a bipartite $|\psi\rangle\equiv\psi_{ij}|ij\rangle$, then measuring in the computational basis the first system you get the state
$$|\psi_i\rangle=\frac{1}{\|(|i\rangle\!\langle i|\otimes I)|\psi\rangle\|}(|i\rangle\!\langle i|\otimes I)|\psi\rangle$$
for some $i$. This obviously contains no correlation between first and second system.
As another example, if you have the three-qubit state $|0\rangle(|00\rangle+|11\rangle)+|1\rangle(|00\rangle-|11\rangle)$, then when the first qubit is found to be $|0\rangle$, the state of the rest of the system is described by $|00\rangle+|11\rangle$.
On the other hand, if you mean to measure the first system and neglect the measurement outcome (which is equivalent to just ignoring the first system, regardless of whether you measure it), then the residual state is to be described via the partial trace. This residual state can be pretty much anything. For example, if the initial state is $|0\rangle\!\langle0|\otimes \frac{I}{2}$, then partial tracing the first system gives the maximally mixed state $\frac{I}{2}$. If the initial state is $|0\rangle\otimes(|00\rangle+|11\rangle)$, then partial tracing the first system gives the maximally entangled residual state.
