We have a structure: $\mathbb Q=\langle \mathbb Q,\,+,\,-,\,*,\,0,\,1 \rangle$
Question: Does $\mathbb Q$ contain a substructure which is not a field?
I have in my notes that there is such a substructure and it is $\mathbb Z$ because $\mathbb Z \subseteq \mathbb Q$
But I don't think that I fully understand it.
We wrote the axioms of field and that
$x \neq 0 \rightarrow (\exists y)(x*y=y*x=1)$ does not satisfy rest of the theory.
but I don't know exactly why.
On
The answer to your question depends on what kind of substructure you're looking for. (That might have been explained in class but didn't get to your notes.)
There is no subset of $\mathbb{Q}$ that is itself a field with the usual addition and multiplication.
$\mathbb{Z}$ is a subring of $\mathbb{Q}$ when you think of the latter as a ring.
Any subset of $\mathbb{Q}$ is a "substructure" if "being a set" is the only structure you care about.
I don't know what "structure" is supposed to mean. But I can easily explain to you why $\mathbb Z$ is not a field.
It is, as you state, because the multiplicative inverse axiom fails. For $\mathbb Z$ to be a field, it must be true that for all $x \ne 0$ in $\mathbb Z$, then exists a $y \in \mathbb Z$ so that $x*y = y*x = 1$. Such a number we would call $y = \frac 1x$. It is called the "multiplicative inverse of $x$" because $x*\frac 1x = \frac 1x *x = 1$.
That simply is not true for $\mathbb Z$. Take any $x \ne \pm 1$. Say for instance $x = 2$. Then there does not exist any integer $y$ so that $x*y = y*x = 1$. (For example: if $2*y = y*2 =1$ then $y = \frac 12$ and $y = \frac 12 \not \in \mathbb Z$).
$\mathbb Z$ is not a field, because it is not true that ever element other than $0$ has a multiplicative inverse. In fact, other than $1$ and $-1$, no integer has a multiplicative inverse.