Consider the morphism of affine varieties $\mathbb{A}^1 \rightarrow \mathbb{V}(y^2 - x^3) \subset \mathbb{A}^2$ defined by $t \mapsto (t^2, t^3).$ Clearly the morphism is a bijection and it also seems clear that we cannot obtain $t$ back from $(t^2, t^3)$ by polynomial maps so the morphism is not an isomorphism. But is there a way to see that this morphism is not an isomorphism visually?
This is a different example. Above is a visualization of $\mathbb{V}(y - x^2)$ and it is easy to see that the morphism $\mathbb{A}^1 \rightarrow \mathbb{V}(y - x^2)$ sending $t \mapsto (t, t^2)$ is an isomorphism as we can project the graph onto $x$ axis.
However, is there a similarly easy way to see that the morphism is not an isomorphism just from this picture above of $\mathbb{V}(y^2 - x^3)$? I guess one thing going for the first example is recognizing that projections are polynomial maps and hence we obtain a straightforward isomorphism.
One of the properties preserved by isomorphism of schemes is whether or not a point is a singularity.
$V(y^2 - x^3)$ has a singular point: the cusp at the origin. $\mathbb{A}^1$ is nonsingular.
So, they can't be isomorphic.