I have come across the following past exam question...
Define an ideal $J:=(z^2x+y^2-2y,x^3+y^3+z^3,x^2+2z^2) \subseteq \mathbb{Q}[x,y,z]. $
Compute a generating set for $J \cap \mathbb{Q}[y]$.
Compute a generating set for $J \cap(y)$.
Compute a Gröbner Basis for $(J:y)$ using lexicographic order with $x>y>z$.
Any help in computing the generating sets and then the Gröbner basis would be greatly appreciated as I'm a bit lost.
This might help you : with the help of my computer I found out that
$$ J \cap {\mathbb Q}[y]=(y^6 + 4y^5 - \frac{7}{2}y^4 - 18y^3 + 18y^2){\mathbb Q}[y] \tag{1} $$
at that the Groebner basis for $(J:y)$ (with respect to the graded lex order with $x \gt y \gt z$) consists of six elements, namely
$$ \begin{eqnarray} h_1 & = & x^2 + 2z^2, \\ h_2 & = & \left( \frac{y}{2} - 1\right)x + (y^2 + 2y - 4)z, \\ h_3 & = & z^3 + (y^3 + 2y^2 - 4y), \\ h_4 & = & z^2x + (y^2 - 2y), \\ h_5 & = & (y^2 + 2y - 4)x + (-y + 2)z, \\ h_6 & = & \left( -\frac{y^2}{2} + y\right)x + z^4 \\ \end{eqnarray} $$
Perhaps there is a reasonable hand-made computation that leads to (2), but it seems unlikely that (1) can be reached without a computer. So there might be a typo in your original question.
Regarding the second question, number your original generators : $$ g_1=z^2x+y^2-2y, \ \ g_2=x^3+y^3+z^3,\ \ g_3=x^2+2z^2\tag{3} $$
Let $f\in J\cap {\bf\big (}y {\bf\big )}$. Then $f$ can be written $f=p_1g_1+p_2g_2+p_3g_3$, where each $p_i$ is a polynomial in $x,y,z$. For each $i$ we can write $p_i=q_i+r_iy$ where $q_i,r_i$ are polynomials and $q_i$ does not contain a $y$ (in fact, $q_i(x,z)=p_i(x,0,z)$). Then the hypothesis $f \in {\bf\big (}y {\bf\big )}$ yields $p_1{g^{*}}_1+p_2{g^{*}}_2+p_3{g^{*}}_3=0$ where for each $i$ we put ${g^{*}}_i(x,z)=g_i(x,0,z)$.
So we are naturally lead to consider the following ideal $I_1$ of ${\mathbb Q}[x,z]^2$ :
$$ I_1=\lbrace (q_1,q_2,q_3) \in {\mathbb Q}[x,z]^2 \ | \ q_1{g^*}_1+q_2{g^*}_2+q_3{g^*}_3=0 \rbrace \tag{4} $$
To study $I_1$, we first look at the projection
$$ I_2=\lbrace q_2 | (q_1,q_2,q_3) \in I_1\rbrace \tag{5} $$
First, we have $x\in I_2$ because
$$ (2x-z)(z^2x)+x(x^3+z^3)+(-x^2)(x^2+2z^2)=0 \tag{6} $$
and we also have $z\in I_2$ because
$$ (\frac{x}{2}+2z)(z^2x)+z(x^3+z^3)+(-\frac{z}{2})(x^2+2z^2)=0 \tag{7} $$
and as it is easy to see that $1\not\in I_2$, we deduce that $I_2={\bf\big (}x,z{\bf\big )}$ and hence that $I_1$ is generated as an ideal by $(2x-z,x,-x^2),(\frac{x}{2}+2z,z,-\frac{z}{2})$ and $(-{g^*}_3,0,{g^*}_1)$.
Plugging this result back into the decomposition of $f$, we deduce the following set of six generators for $J\cap {\bf\big (}y {\bf\big )}$ :
$$ yg_1,yg_2,yg_3, (2x-z)g_1+xg_2+(-x^2)g_3, (\frac{x}{2}+2z)g_1+zg_2+(-\frac{z}{2})g_3, (-{g^*}_3)g_1+{g^*}_1g_3. $$
