The set $\mathbb{M}$ of all monomials forms a semigroup, and the set of leading monomials $LM(I)$ is a semigroup ideal in $\mathbb{M}$. The ideal basis $G=(g_1,\ldots,g_s)$ is a Gröbner basis of $I=\langle g_1,\ldots,g_s\rangle$ iff $LM(I)$ is generated as a semigroup ideal by $LM(g_1), \ldots, LM(g_s)$.
The two definitions are equivalent.
Just for clarity, the exercise is in section 2.5 of the book.
The inclusion $\langle LT(g_1),\ldots,LT(g_t)\rangle \subset \langle LT(I) \rangle$ always holds, as $\{g_1,\ldots,g_t\} \subset I$ and $\langle LT(I) \rangle$ is the ideal generated by elements of $LT(I) = \{LT(f) \mid f \in I, f \neq 0\}$. You have correctly identified the leading terms in the left-hand side. The question is whether there exists nonzero $f \in I$ such that $LT(f) \not\in \langle LT(g_1),\ldots,LT(g_t) \rangle$.
That is, can you make a linear combination (with polynomial coefficients) of $g_1,\ldots,g_t$ such that the result has its leading term not in $\langle LT(g_1),\ldots,LT(g_t) \rangle$? A natural idea is to make a linear combination such that the "would-be" leading term (looking at the leading terms of the summands) would cancel, so that you get a "possibly new" leading term. In this particular case we have e.g. $LT(g_1) = x^4y^2$ and $LT(g_2) = x^3y^3$, so we can multiply $g_1$ by $y$ and $g_2$ by $x$ to get the same leading terms $LT(yg_1) = x^4y^3$ and $LT(xg_2) = x^4y^3$, and now we consider the linear combination $f=yg_1 - xg_2 \in I$. In fact $f=yg_1 - xg_2 = y(x^4y^2 - z^5) - x(x^3y^3-1) = -yz^5 + x$, so its leading term is $LT(f) = -yz^5$, and this is not contained in the ideal $\langle x^4y^2, x^3y^3,x^2y^4 \rangle$. Recall that checking membership of a monomial in a monomial ideal is easy; see Lemma 2 of section 2.4 in the book. So we have found a nonzero $f \in I$ such that $LT(f) \not\in \langle LT(g_1),LT(g_2),LT(g_3) \rangle$, and hence $\{g_1,g_2,g_3\}$ is not a Groebner basis.
This heuristic idea leads to the definition of the S-polynomial later in the book (the next paragraph 2.6), and from there to Buchberger's criterion and Buchberger's algorithm.
Best Answer
It suffices to pick a nonzero $f\in I$ and show that $LT(f)\in\langle LT(g_1),\ldots LT(g_s)\rangle$. By properties of monomial ideals, we just need to show that $LT(f)$ is divisible by one of the $LT(g_i)$. Since $f$ has a remainder of zero upon division by $G$, then it must be the case that $LT(f)$ is divisible by one of the $LT(g_i)$ just by using the definition of the division algorithm. Otherwise, if $LT(f)$ weren't divisible by any $LT(g_i)$, then $LT(f)$ would show up as a nonzero term in the remainder (but this cannot happen since the remainder was assumed to be zero).