Let $k$ be an algebraically closed field and consider the closed subschemes $X=\operatorname{Spec}(k[x,y,z]/(x-yz))$ and $Y=\operatorname{Spec}(k[x,y,z]/(x^2-yz))$ of $\mathbb{A}^3$. Then the gradient of $x-yz$ at the origin $O$ is $(1,0,0)$, so $X$ is smooth at $O$, while the gradient of $x^2-yz$ at the origin is $(0,0,0)$, so $Y$ isn't smooth at $O$.
I don't understand the intuitive reason why $X$ is smooth at $O$ while $Y$ isn't. After all, $X$ can be seen as a family of hyperbolae $yz=x$ varying over $\mathbb{A}^1_x$, and so over the origin we get an intersection of two lines. I understand that (non)smoothness of a closed subscheme is in some sense independent of the ambient scheme (e.g. a closed subscheme can by singular at a point where the ambient scheme is smooth), but nevertheless I have a hard time thinking of the total scheme $X$ as something smooth, because near the origin it is "self-intersecting".
And then what exactly is it that makes $Y$ singular at the origin? I.e. why doesn't a possible intuitive explanation of why $X$ is smooth apply to $Y$ aswell?
I'm sorry if the question is vague, but I'm hoping to understand the notion of smoothness in a more intuitive manner.
The quotient ring $k[x,y,z](x-yz)$ enforces $x$ to become the product of $y$ and $z$, which makes it obsolete. Therefore this ring is just isomorphic to $k[y,z]$ and you end up with an affine space. This should clarify why the first scheme is smooth.
The second equation (which btw does not make $x$ obsolete since it only tells you something about $x^2$) results in a scheme whose real image is given by two cones intersecting at their tips, a point where tangent vectors cannot be defined properly.
Another kind of intuition can come from using the etale topology, which is closer to the analytic one (so visualize the analytic one). Here the relevant statement is that smooth schemes are etale-locally affine spaces (of the right dimension, of course). Certainly this is not given at this point if we visualize the scheme as before.