Showing that an algebraic set is not isomorphic to $\mathbb{A}^1$ (For convenience, I'm assuming $\mathbb{K} = \mathbb {C}$.) 

I'm trying to show that the algebraic set $\mathbb{V}(xz-y^2,x^3-yz,z^2-x^2y) \subseteq \mathbb{A}^3$ is not isomorphic to $\mathbb{A}^1$. 

I already know that $X = \{(t^3,t^4,t^5) \in \mathbb{A}^3 \;|\; t \in \mathbb{A}^1 \} $, and that $\mathbb{I}(X) = \langle xz-y^2,x^3-yz,z^2-x^2y \rangle$.
If there were an isomorphism between the $\mathbb{A}^1$ and $X$, then $\mathbb{K}[\mathbb{A}^1] \cong \mathbb{K}[X]$. But $\mathbb{K}[\mathbb{A}^1] \cong \mathbb{K}[t]$, whereas $\mathbb{K}[X] \cong \mathbb{K}[t^3,t^4,t^5]$ by considering the homomorphism 
$$\mathbb{K}[x,y,z] \rightarrow \mathbb{K}[t], \; x\rightarrow t^3, \; y\rightarrow t^4, \; z\rightarrow t^5$$
which has kernel $\mathbb{I}(X)$ and image $\mathbb{K}[t^3,t^4,t^5]$, so 
$$\mathbb{K}[X] \cong \mathbb{K}[x,y,z]/\mathbb{I}(X) \cong \mathbb{K}[t^3,t^4,t^5]$$
Thus we would have that $\mathbb{K}[t] \cong \mathbb{K}[t^3,t^4,t^5]$, but this is not the case since the former is a UFD, whilst the latter is not, since for example $t^8 = t^3t^5 = (t^4)^2$ has two factorisations. Hence, $\mathbb{A}^1 \ncong X$.
Is this line of reasoning correct? Is there an easier way I could have gone about it?
 A: Alternative way is to observe that $X$ is not smooth at the origin. In other words, localization of $\mathbb{C}[x,y,z]/I$ at $\mathfrak{m}_0=\langle x,y,z\rangle$ is not regular, where $I=\langle xz−y^2,x^3−yz,z^2−x^2y\rangle$. But $\mathbb{C}[t]$ is smooth everywhere. 
Let us verify the above claim. We can use the Jacobian criteria to test regularity (see for instance Eisenbud's Commutative Algebra, Chapter 16):
Suppose $k$ has char. $0$, and $I=(f_1,\dots,f_m)$ is an ideal in $S:=k[x_1,\dots,x_n]$. Suppose $\mathfrak{p}\supset I$ is prime and $I_\mathfrak{p}$ has codimension $r$ in $S_\mathfrak{p}$. Define the Jacobian matrix as 
$$
J_{ij}=\frac{\partial{f_i}}{\partial x_j}
$$
Then $(S/I)_\mathfrak{p}$ is a regular local ring if and only if $J$ has rank $r$ in $S/\mathfrak{p}$. 
(This is saying that regularity and smoothness are the same things in char. $0.) 
Now in our case, we note that $I_{\mathfrak{m}_0}$ has codimension $>0$ in $\mathbb{C}[x,y,z]_{\mathfrak{m}_0}$ (in fact it has codimension $3$, but we don't need this). But the Jacobian
$$
J=\left(\begin{array}{ccc}z& -2y&x\\3x^2&-z&-y\\ -2xy&-x^2&2z\end{array}\right)
$$
has rank zero in $\mathbb{C}[x,y,z]/\mathfrak{m}_0$. Thus the origin is not a smooth point of $X$.
