How to show that if x is a cluster point of a filter then it a cluster point of each of its associated or derived net. Am stuck with this problem. However I am able to show the other way round. Please help me wih it.
 A: Typically the associated net to a filter $\mathcal{F}$ on $X$ has as its index set
$$I= \{(F,x) \in \mathcal{F} \times X\mid x \in F\}$$ with order
$$(F_1,x) \le_I (F_2,x') \iff F_2 \subseteq F_1$$
which is directed as a filter is closed under finite intersections.
The net $n: I \to X$ is then simply defined as the projection $n((F,x))=x$.
(maybe the OP allows more of such associated nets on just the index set of reverse-inclusion $\mathcal{F}$, but then choice is required to get such a net; IMO this is more "canonical").
Suppose $p$ is a cluster point of $\mathcal{F}$ which means
$$\forall O \in \mathcal{N}_p: \forall F \in \mathcal{F}: F \cap O \neq \emptyset\tag{1}$$
(or equivalently, $p \in \bigcap \{\overline{F}\mid F \in \mathcal{F}\}$).
Now we will show $p$ is also a cluster point of the net $n$: let $O$ be any neighbourhood of $p$ and $i_1 = (F_1,x)$ be any member of $I$, we must find a larger member $i_2$ of $I$ such that $n(i_2) \in O$ (the net must be frequently in all neighbourhoods of $p$).
Applying $(1)$ to $O$ and $F_1$ we find $y \in O \cap F_1$. But then trivially $i_2 = (F_1,y) \in I$ and also $i_1 \le_I i_2$ and $n(i_2)= y \in O$. So we are already done.
It's basically almost "true by definitions"; nothing deep.
Added after comments
If any net $n$ on $\mathcal{F}$ (ordered by reverse inclusion) to $X$, such that $n(F) \in F$ for all $\mathcal{F}$, counts as an associated net of $\mathcal{F}$, the statement must be adapted I think. It's not true (see example by OP in a comment) that for all associated nets to $\mathscr{F}$, $p$ must be a cluster point of $n$, but we maybe could say that this is the case for some associated net. The "canonical" associated net above does work, but we cannot claim for all in the general definition.
So in that case the general statement from the title of the question is false.
