Consider the function $\displaystyle f(z)=\frac{1}{\cos\frac{1}{z}}$. Test what kind of singularity at $z=0$ ?
For singularities of $f(z)$ , $\cos \frac{1}{z}=0\implies z=\frac{2}{\pi (2n+1)}$. Now $z=0$ is also a singularity of $f$ and $0$ is the limit point of poles.
$\bullet$ So from a known theorem (which stated as , "Limit point of poles is an essential singularity") , $0$ is an essential singularity of $f$.
$\bullet$ Definition : (from the book : Functions of One Complex Variables by J.B. Conway) An isolated singularity of $f$ is said to be an essential singularity of $f$ if it is neither a pole nor a removable singularity of $f$.
$\bullet$ Again , from the singularities of $f$ it is clear that $0$ is a non-isolated singularity , as in any neighbourhood of $0$ there is also a singularity of $f$ other than $0$.
$\bullet$ Thus we get , $0$ is a non-isolated essential singularity of $f$. Which violates the definition of the Essential singularity.
Where my fallacy ? Please detect and rectify..
Best Answer
Use the characterization based on limits. A removable singularity has a limit at the given point. A pole happens when the function is unbounded near the point. An essential singularity happens when neither come to pass. So at $0$ since $\displaystyle\lim_{x\to\infty} \cos(x)$ does not exist, clearly that singularity is essential since it does not converge to a limit, and it is not unbounded on the real axis.
Written after the question was edited
Your fallacy is using the word "isolated" when you don't really mean that: in you case it seems you mean the adjective "essential" to only be applicable to isolated singularities. However, this seems to be in direct conflict with this "known theorem" you allude do, since clearly a limit of singularities can never be isolated. As such, I think the most reasonable interpretation of your question falls into the original answer I posted, but with the definition of "essential singularity" you give, there is no terminology defined for a "non-isolated essential singularity."
In short: your question cannot be answered as-stated, but this "known theorem" seems to indicate it's most likely your class admits non-isolated essential singularities as a thing.