The confusion may rely on your definition of $\cong$.
If $\cong$ means group isomorphism then the formula $(2\mathbb Z/6\mathbb Z) \cong (\mathbb Z/3\mathbb Z)$ is true.
If the use of $\cong$ here means canonically isomorphic then the formula is false.
Recall that for a given class of algebraic structures
(say rings, groups, fields, ...) two objects in the class
are said to be canonically isomorphic if there is a unique
isomorphism between them.
In your examples $3\mathbb Z/6\mathbb Z$ is a group of two elements and admits a unique
isomorphism with $\mathbb Z/2\mathbb Z$, so they are canonically isomorphic.
Indeed, in an isomorphism
3\mathbb Z/6\mathbb Z\rightarrow \mathbb Z/2\mathbb Z
the class of $0$ needs to be mapped to the class of $0$. This leaves no freedom of choice for the image of $$.
However, in the case $2\mathbb Z/6\mathbb Z\rightarrow \mathbb Z/3\mathbb Z$ there
are $2$ isomorphisms (of groups) completely determined by the image of $2$. In this case any chocie $\mapsto $ or $\mapsto $
defines an isomorphism.
Let $m,k$ be positive integers.
Then $k \mathbb Z/mk\mathbb Z$ is a cyclic group of order $m$. Thus, it is isomorphic to $\mathbb Z/m\mathbb Z$ as groups. The number of possible isomorphisms $k \mathbb Z/mk\mathbb Z\rightarrow \mathbb Z/m\mathbb Z$ coincides with the number $\varphi(m)$ of generators of $\mathbb Z/m\mathbb Z$. Here $\varphi(m)$
denotes the Euler's totient function which is $\geq 2$ if $m \geq 3$.
Hence $k \mathbb Z/mk\mathbb Z\cong \mathbb Z/m\mathbb Z$ is always true as group
isomorphism. However $k \mathbb Z/mk\mathbb Z$, $\mathbb Z/m\mathbb Z$
are only canonically isomorphic when $m=1$ or $2$.
Notice that the canonically part depends strongly on the class of structures you consider, as the comments suggest. The ring $2\mathbb Z/6\mathbb Z$ is unital (the unit is $4$) and $2\mathbb Z/6\mathbb Z$ is canonically isomorphic to $\mathbb Z/3\mathbb Z$ as unital rings.