Let $W_1 = V(\langle u\rangle), W_2 = V(\langle v^2-u^3\rangle) \subset \mathbb{A}^2(\mathbb{C})$, $V_1 = V(\langle x, y\rangle) \subset \mathbb{A}^3(\mathbb{C})$ and $V_2 = V(\langle z, y^2-x^3\rangle) \subset \mathbb{A}^3(\mathbb{C})$ be affine algebraic varieties.
It seems intuitively obvious that the pairs of varieties $V_1, W_1$ and $V_2, W_2$ are isomorphic, but how can I show that? Here's my attempt (for $V_2, W_2$):
Define $\phi : W_2 \to V_2$ by the function $\phi(u,v) = (u, v, 0)$. Then clearly $u,v,0\in \mathbb{C}[W_2]$, so $\phi$ is indeed a morphism $\phi : W_2 \to \mathbb{A}^3(\mathbb{C})$, and we know that $u$ and $v$ must satisfy $v^2 - u^3$. But furthermore the image of $(u,v)$ is contained within the variety $V_2$. Therefore $\phi$ is a morphism $\phi : W_2 \to V_2$. Consider the function $\psi : V_2 \to W_2$ defined by $\psi(x,y,0)=(x,y)$, which is easily seen to be an inverse morphism for $\phi$ whose image is contained in $W_2$. Therefore we have found an explicit isomorphism of varieties, so the varieties are isomorphic.