Q[x,y]/(x) = Q[y]? [closed]

For any commutative ring $\,R\,$ we have $\,R[x]/(x) \cong R.\,$ Indeed the evaluation hom $\,R[x]\to R\,$ via $\,x\mapsto 0\,$ has kernel $\,(x),\,$ since $\,f(0)=0\iff x\mid f(x)\,$ by the Factor Theorem.

Yours is the special case $\, R = Q[y],\,$ since $,Q[x,y]\cong (Q[y])[x].$

The easiest way, probably, is to define $f:R[x,y]\rightarrow R[y]$ by $f|_{R[y]} = \text{id}_{R[y]}$ and $x\mapsto 0$. You can check that this is a homomorphism. Next we check that the kernel of this map is $(x)$, so that our result follows by the first isomorphism theorem. Note that $f(x) = 0$, so $(x)\subseteq\ker(f)$. Next, choose $\alpha\in R[x,y]$ and suppose $f(\alpha) = 0$. Define $\tilde{\alpha}$ as the polynomial $\alpha$ with all factors that contain $x$ removed. Then $\tilde{\alpha}\in R[y]$ and $f(\alpha) = f(\tilde\alpha) = \tilde\alpha$. Thus $f(\alpha)=0$ iff $\tilde\alpha=0$ iff $\alpha\in(x)$. We conclude that $\ker(f)\subseteq(x)$, which concludes the statement.