This kind of formula only works when you have the action of the lie group $G = SO(3)$ on some manifold $M$, in this case, the euclidean space $\mathbb{R}^3$. Let then $G$ act on $M$ by $(g,x) \mapsto gx$.
A left-invariant vector field $V$ in $G$ induces a vector field on $M$ such that the flow in $M$ is: $${\gamma(t,x) = \exp(tV)x,}$$ where $\exp: \mathfrak{g} \rightarrow G$ is the exponential map. Define the induced vector field $\tilde{V}$ on $M$ by: $${\tilde{V} = \frac{d}{dt}\exp(tV)x|_{t=0}.}$$Then we have defined a map $\mathfrak{g} \rightarrow \mathfrak{X}(M)$ by $V \mapsto \tilde{V}$.
In your case, take an element of $\mathfrak{so}(3)$ (the lie algebra of $SO(3)$, e.g.: the tangent space at the identity), namely:
$$X = \begin{bmatrix}
0 & 0 & 0\\
0 & 0 & -1\\
0 & 1 & 0
\end{bmatrix}$$
(You can see that this is an element of $\mathfrak{so}(3)$ because it is an antissymetric matrix).
Then your induced vector field is $\tilde{X} = -z\frac{\partial}{\partial y} + y \frac{\partial}{\partial x}$.
Recall that any Riemannian manifold $(M,g)$ admits a unique symmetric metric connection $\nabla$ by the fundamental theorem of Riemannian geometry, and it is determined by the Koszul formula:
$$2g(\nabla_Y X, Z) =Xg(Y, Z) + Yg(X, Z) - Zg(X, Y) -g([X, Z], Y) - g([Y, Z], X)
- g([X, Y], Z)$$
Suppose now that we had a compact Lie group $G$ with a bi-invariant metric $\eta$. There's a special class $\mathcal{G}$ of vector fields on $G$ which are left-invariant, i.e., $X \in \mathcal{G}$ iff $dL_g X_h = X_{gh}$ where $L_g : G \to G$ is multiplication by $g$. Any element $X$ of $\mathcal{G}$ is obtained from left-translating a tangent vector $X_e$ in $\mathfrak{g}$ to extend to all of $G$. Notice that left-invariant vector fields have constant inner product (or angles between them stay globally constant) because for any two left-invariant vector fields $U$ and $V$, $$\eta(U_{gh}, V_{gh}) = \eta(dL_gU_h, dL_g V_h) = \eta(U_h, V_h)$$ where the second equality derives from the left-invariance of $\eta$.
Therefore if $X, Y, Z$ are vector fields in $\mathcal{G}$, $\nabla$ a symmetric connection on $G$ compatible with $\eta$, we invoke the Koszul formula. By the previous comment, $X\eta(Y, Z), Y\eta(X, Z)$ and $Z\eta(X, Y)$ are all zero so we are left with
$$2\eta(\nabla_Y X, Z) = -\eta([X, Z], Y) - \eta([Y, Z], X) - \eta([X, Y], Z)$$
Up until this point we have not needed the right-invariance of $\eta$. But the following lemma is the relevant key: If $U, V, W$ are vectors fields in $\mathcal{G}$, then $\eta([U, V], W) = \eta(U, [V, W])$. To see this, let $g_t$ be the element of $G$ defined as $\partial_t g_t = V(g_t)$ with initial condition $g_0 = e$ (so it's a flowline of $V$ starting at $e$). Write $\eta(U, W) = \eta(dR_{g_t}U, dR_{g_t}W)$ using right-invariance of $\eta$. Differentiate with respect to $t$ at $t = 0$ to obtain $0 = \eta(\mathcal{L}_V U, W)+\eta(U, \mathcal{L}_V W)$, and writing the Lie derivatives as brackets gets the desired formula.
Therefore $\eta([X, Z], Y) = -\eta([X, Y], Z)$ and $\eta([Y, Z], X) = \eta([X, Y], Z)$, which upon plugging in the previous formula we obtain $2\eta(\nabla_Y X, Z) = -\eta([X, Y], Z)$. Since this holds for all left-invariant vector fields $Z$, we conclude the formula (after using $[X, Y] = -[Y, X]$)
$$\boxed{2\nabla_Y X = [Y, X] = \mathcal{L}_Y X}$$
In particular your question is easily answered, as $\nabla_X X = 1/2 \mathcal{L}_X X = 0$, implying every left-invariant vector field $X$ on $(G, \eta)$ is parallel.
One can completely understand the geodesics of $G$ using this fact. Consider the Riemannian exponential map $\text{exp} : \mathfrak{g} \to G$ given by $\exp(v) = \gamma(1)$ where $\gamma$ is the solution to the ordinary differential equation $\nabla_{\gamma'} \gamma' = 0$ with initial condition $\gamma(0) = e$, $\gamma'(0) = v$. If we define the left-invariant vector field $V$ on $G$ as $V_g = dL_g v_e$, then consider the integral curve $\gamma_V$ of $V$ defined as the solution to the ordinary differential equation $\gamma_V'(t) = V(\gamma_V(t))$ with initial condition $\gamma_V(0) = e$. Definitionally, the tangent field to $\gamma_V$ is $V$, and $\nabla_V V = 0$ as $V$ is left-invariant, so $\gamma_V$ is a solution to the initial value problem, therefore by uniqueness $\gamma_V = \gamma$.
So alternatively we can define $\exp : \mathfrak{g} \to G$ as $\exp(v) = \gamma(1)$ where $\gamma$ is a one-parameter subgroup of $G$ with $\gamma(0) = e$ and $\gamma'(0) = v$ ($\gamma$ is in fact tangent to $V = dL v$). The geodesics of $G$ are exactly the image of one-parameter subgroup of $G$.
As an addendum, here's one argument to construct a bi-invariant metric on any compact connected Lie group $G$ (a hands-on construction is given in doCarmo, "Riemannian Geometry" problem $7$ of Ch. $1$). Pick any Riemannian metric $g(-, -)$ on $G$, and let $\mu$ be the Haar measure on $G$ (which is a bi-invariant measure). Define $\eta$ by the averaging formula
$$\eta(X, Y) = \int_{G \times G} g(dL_g dR_h X, dL_g dR_h Y)\; d\mu(g) d\mu(h)$$
Clearly $\eta$ is bi-invariant by construction (this argument is in Milnor, "Morse theory", Ch. $21$)
Best Answer
You could start taking the Lie group $(\mathbb R^n,+),$ and considering what does it means for a vector field $X$ on $\mathbb R^n$ to be left-invariant.