I would like to show that the following algebraic sets:
$V_1 = Z(xy,yz,xz) \subset \mathbb{A}^3$ and $V_2 = Z(xy(x+y),z) \subset \mathbb{A}^3$
are not isomorphic.
What I have so far: All I know is that it suffices to prove that the coordinate rings are not isomorphic
that is - $\mathbb{C}[x,y,z]/(xy,yz,xz) = \mathbb{C}[V_1] \ncong \mathbb{C}[V_2] = \mathbb{C}[x,y,z]/(xy(x+y),z)$
as $\mathbb{C}$-algebras.
Thanks in advance!
Let's expand on the hint from the comments. You've successfully identified that the two sets are both the union of three lines - one is the union of 3 axes in $\Bbb A^3$, and the other is three lines in a plane (the two axes and the line $x+y=0$). We can for sure define a map from $V_1$ to $V_2$ by sending the $x$ and $y$ axis to themselves and squashing the $z$-axis in to the plane in the direction of that third line, but intuitively, this should not be reversible. Something about that squashing shouldn't be able to be undone.
To prove this, we'll show that the (co)tangent spaces at the unique point where all three lines meet are different dimensions. If you haven't learned about this yet, don't worry - there's a ring-theoretic explanation that won't cause you too much trouble. We'll be computing $\mathfrak{m}/\mathfrak{m}^2$ for all the maximal ideals in $\Bbb C[V_1]$ and $\Bbb C[V_2]$.
On an axis away from the origin, the maximal ideal is of the form $(x-a,y,z)/I(V_i)$ with $a\in\Bbb C^*$ (up to permutation of $x,y,z$), and it's not to bad to see from direct calculation that $\mathfrak{m}/\mathfrak{m}^2$ is a dimension one vector space on the basis element $x-a$. On the line cut out by $(x+y)$, we have to think a little harder, but not really: up to a linear change of coordinates, we can just move over to an axis and it's all fine. So we've shown that $\mathfrak{m}/\mathfrak{m}^2$ is a dimension one vector space for every maximal ideal $\mathfrak{m}$ in $\Bbb C[V_1]$ and $\Bbb C[V_2]$ that doesn't correspond to the origin.
Now let's look at the origins. In $\Bbb C[V_1]$, the maximal ideal of the origin is $(x,y,z)/(xy,yz,zx)$, so it's square is $(x^2,xy,y^2,xz,yz,z^2)/(xy,yz,zx)=(x^2,y^2,z^2)/(xy,yz,zx)$, and we can see in the quotient $\mathfrak{m}/\mathfrak{m}^2$ we have three surviving linearly independent terms: $x,y,z$. In $\Bbb C[V_2]$, the maximal ideal of the origin is $(x,y,z)/(xy(x+y),z)$ and it's square is $(x^2,xy,y^2,xz,yz,z^2)/(xy(x+y),z)=(x^2,xy,y^2)/(xy(x+y),z)$. The quotient only has two surviving linearly independent terms: $x,y$. So $\Bbb C[V_1]$ has maximal ideals who's quotient by their squares are dimensions 1 and 3 (but not 2) as vector spaces, and $\Bbb C[V_2]$ has maximal ideals who's quotients by their squares are dimensions 1 and 2 (but not 3). Since any isomorphism between the two algebras would preserve the dimensions of the corresponding quotients of maximal ideals by their squares, we see the two algebras cannnot be isomorphic.
(Alternatively, one could note that $\Bbb C[V_1]$ requires 3 generators as an algebra, while $\Bbb C[V_2]$ only needs 2. But the above is fun too!)