I know how to prove, for example, that $\mathbb{Q}[x]/(x) \cong \mathbb{Q}.$ You consider the homomorphism $f:p(x) \mapsto p(0)$. It is clear that $(x) \subseteq \ker f$, and then you can use the Divison Algorithm to show that $\ker f = (x)$, and then use the First Isomorphism Theorem.
I believe I have heard people claim the following facts as well: $$1)\ \mathbb{Q}[x, y, z]/(x) \cong \mathbb{Q}[y, z]$$ $$2)\ \mathbb{Q}[x, y, z]/(x, y) \cong \mathbb{Q}[z]$$
I would think you do them the same way. For $2)$, we would use the homomorphism $f:p(x, y, z) \mapsto p(0, 0, z)$. However, I run into two issues:
1) How do we prove that $f$ is really a homomorphism? For univariate polynomials, we can make use of the explicit formulas for multiplication and addition. For example, the multiplication formula is: $$\left( \sum_{k=0}^n a_k x^k \right) \left( \sum_{k=0}^m b_k x^k \right) = \sum_{k=0}^{n+m} \left( \sum_{i=0}^k a_k b_{i-k}x^k\right) $$ (where $a_i$ and $b_i$ are define as zero for $i$ greater than their degree). For multivariable polynomials, I have never seen the explicit formula, but I assume it's a complete nightmare.
2) Sticking with example 2, how do we prove that the kernel of $f:p(x, y, z) \mapsto p(0, 0, z)$ is $(x, y)$? It is easy to see that the kernel contains $(x, y)$. In the univariate case we used the Division Algorithm to prove the reverse containment, but as far as I know, multivariate polynomials rings are, in general, not Euclidean Domains.
Thank you for your help.