Is this Lie group isometric to the Euclidean plane? Consider the Lie group
$$G := \left\{ \begin{pmatrix}
1 & 0 & 0 \\
x & 1 & 0 \\
y & x & 1
\end{pmatrix}: x, y \in \mathbb{R}\right\}$$
equipped with the left-invariant Riemannian metric $g$ given by the left translation in $G$. Then, $g$ has expression
$$ g = (1 + x^{2}) dx \otimes dx - x dx \otimes dy - x dy \otimes dx + dy \otimes dy.$$
Easily, we can compute the Christoffel symbols of the Levi-Civita connection. The only non-zero symbol is $\Gamma_{11}^{2} = -1$. Then, $G$ has constant sectional curvature $K = 0$, as the Euclidean plane $\mathbb{R}^{2}$. Now, are $\mathbb{R}^{2}$ and $G$ isometric Riemannian manifolds? I know that if two Riemannian manifolds have  different sectional curvatures in corresponding points, they cannot be isometric but, this not the case. Should I try to find conditions for the existence of such isometry?
Thanks in advance.
SOLUTION:
As @Didier pointed out, one way to prove that $\mathbb{R}^{2}$ and $G$ are isometric is to use that $G$ is a complete simply connected Riemannian manifold with constant curvature $K = 0$, so by Theorem 4. 1 (page 163) of Do Carmo's book Riemannian Geometry, $\mathbb{R}^{2}$ and $G$ are isometric. Nevertheless, we can also find explicitly an isometry $\phi: \mathbb{R}^{2} \to G$. Note that
$$\phi(x,y) = \begin{pmatrix}
1 & 0 & 0 \\
\phi_{1}(x,y) & 1 & 0 \\
\phi_{2}(x,y) & \phi_{1}(x,y) & 1
\end{pmatrix},$$
where $\phi_{j}: \mathbb{R}^{2} \to \mathbb{R}$ is differentiable. Let's say that $A$ is the matrix of the metric $g$ of $G$, so we want to get a matrix $B$ (the matrix of the differential of $\phi$), such that $B^{T} A B = I$. Doing Gaussian operations in the matrix $A$ we find that
$$B = \begin{pmatrix} 1 & 0 \\
x &1 \end{pmatrix},$$
so $\phi_{1}(x, y) = x$ and $\phi_{2}(x,y) = \frac{x^{2}}{2} + y$.
 A: The solution edited into the original post is brilliant. I'll just type a different way in which one could have found it (or rather: found the explicit isometry between $G$ and the plane). But in a sense I am cheating because I could only understand how this situation works when I read your solution. Still it can be nice for future readers to see different paths to the same solution. And writing it down is also a test for myself if I really understood it.
Solution:
You may have noticed that your map $\phi$ is not just an isometry for the invariant metric, but also a Lie-group isomorphism. Now if we somehow knew on forehand that all isometries are also isomorphisms and vice versa this would (in my view, but I like algebra better than geometry, which may not be true of everyone) make it a lot simpler to find this $\phi$, or another example that works.
But do we know that isomometries are isomorphisms? My claim is we do.
Isometries between $\mathbb{R}^2$ and $G$ are isomoprhisms:
Given a isometry $\phi$ from the affine plane to $G$ we turn the affine plane into a linear plane and hence into a group by choosing the origin to be the $\phi^{-1}(I)$ with $I$ the identity matrix in $G$.
The fact that the metric is left-invariant means that for each $g \in G$, the map $L_g \colon h \mapsto gh$ is an isometry of $G$. This means that the set $L_G = \{L_g \colon g \in G\}$ is a subgroup of the isometry group of the plane, isomorphic to $G$. Hence $L_G$ is abelian and two-dimensional. The isometry group of the plane is an extremely familiar object. It consists of relections, rotations, translations and glide-reflections that interact in ways that you can almost entirely picture inside your head. From there it is not hard to convince yourself that the only two-dimensional abelian subgroup is the group of translations.
Now this means that for each $a, b \in \mathbb{R}^2$ and any isometry $\phi \colon \mathbb{R}^2 \to G$ sending $0$ to $I$ there exists an element $c$ depending only on $a$ such that
$$\phi(a)\phi(b) = \phi(c + b).$$
(Think as $b$ as a point and $\phi(a)$ as an element of the group $G$. The isometry $L_{\phi(a)}$ of $G$ must correspond under $\phi$ to a translation in the plane by some element $c$.)
By taking $b = 0$ we find from $\phi(0) = I$ that $c = a$. But then the above equation turns into the statement that $\phi$ is an isomoprhism.
Lie group isomoprhisms between $\mathbb{R}^2$ and $G$ are isometries:
Starting with a Lie algebra isomoprhism $\phi$ we can just define a metric on $G$ for which $\phi$ is an isometry by pushing forward the standard metric on $\mathbb{R}^2$. Then we use the isomorphism property to show that this new metric on $G$ is left-invariant. We conclude that it must be equal to the metric from the original post (appealing to the relevant uniqueness theorem) and hence that $\phi$ was an isometry for that metric too.
I find this easier to do with the global metric, i.e. the distance between two points, than with the infinitesimal one. We define the distance $d(x, y)$ between two point $x, y \in G$ as $\|\phi^{-1}(x) - \phi^{-1}(y)\|$ owing to how by Pythagoras' Theorem we have that the distance between two points $a, b \in \mathbb{R}^2$ equals $\|a - b\|$.
Now since $\phi^{-1}$ is an isomorphims, the thing we take the norm of equals $\phi^{-1}(x^{-1}y)$ and so we conclude:
$$d(x, y) = \|\phi^{-1}(x^{-1}y)\|$$
To see that this $d$ is invariant under left-multiplication by $G$, take $g \in G$. We find:
$$d(gx, gy) = \|\phi^{-1}((gx)^{-1}(gy))\| = \|\phi^{-1}(xg^{-1}gy)\| = \|\phi^{-1}(x^{-1}y)\| = d(x, y)$$
as promised.
Follow-up question:
My proof that isometries are isomorphisms and vice versa relies on some explicit properties of the plane. So the question is: in what generality does it hold for general Lie groups $G$ and $H$ both equiped by their invariant metric, that origin-preserving isometries between them are isomorphisms and vice versa? I don't know the answer to this. I think an equivalent (or otherwise closely related) formulation of the question is:
Which Lie groups are uniquely determined by their invariant metric?
