= Solution
A smooth <vector field> on a <smooth manifold> $M$ is a smooth section of its <tangent bundle>: it assigns $v(p)\in T_pM$ smoothly to each point. On a <Lie group>, it is a <left-invariant vector field> when
$$
v(gp)=(dL_g)_p v(p),\qquad L_g(p)=gp.
$$
For $X\in T_eG$, define $v_X(g)=(dL_g)_eX$. Smoothness of multiplication makes this a smooth <vector field>, and $L_gL_p=L_{gp}$ plus the chain rule proves left invariance. Every <left-invariant vector field> arises this way from its value at the identity.
For a matrix <Lie group>, $v_X(g)=gX$, because the derivative at zero of $g\gamma(t)$ is $gX$. The printed group name in this subpart must be read as $\mathrm{SL}_2$, rather than the additive vector space $\mathfrak{sl}_2$: it is the matrix-group calculation consistent with the preceding subparts. At the specified matrix,
$$
\boxed{v_X\!\left(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\right)
=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}
=\begin{pmatrix}0&0\\0&-1\end{pmatrix}.}
$$
A tangent vector at a nonidentity group element need not itself be trace-free; here $g^{-1}v_X(g)=X$ is trace-free, as required. If one instead interpreted the printed $\mathfrak{sl}_2$ as its additive <Lie group>, the field would be the constant field $X$, a different problem.
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