Showing the following sequents are not valid $\lnot \forall x P(x) \models \forall x \lnot P(x)$
$P(x) : x$ is divisible by 2
or $P(x) : x$ is a dog.
or $P(x) can be anything.

I want to show the following sequents are not valid ..Isn't the following sequents equal or the first sequent the whole negation of the statement and the second sequent is the negation of the $P(x)$ .. Can anybody explain this with examples?
First, $\phi \models \psi $ means that $\psi$ is a consequence of $\phi$. In other words, if $M$ is a structure and if $M \models \phi$ ($M$ models $\phi$) then $M \models \psi$ ($M$ also models $\psi$).

Now, regarding your statement $$\lnot \forall x P(x) \models \forall x \lnot P(x)$$
clearly $\forall x \lnot P(x)$ is not a consequence of $\lnot \forall x P(x)$. To show it consider the set of natural numbers as a structure $M$ and $P(x)$ as the predicate "$x$ is divisible by $2$". Then 
$$\lnot \forall x P(x)$$
means "Not for all $x$, $x$ is divisible by $2$ which is a true sentence (since in fact odd numbers are not divisible by 2), while 
$$ \forall x \lnot P(x)$$ 
means "For all $x$, $x$ is not divisible by $2$" which is not a true sentence (since in fact even numbers are divisible by $2$). 

So, the structure $M$ (the set of natural numbers) is an example showing that $M \models \lnot \forall x P(x)$, but $M \nvDash  \forall x \lnot P(x)$ and your statement 
$$\lnot \forall x P(x) \models \forall x \lnot P(x)$$
is not true.