I have a Lie group $\big(G,\,\circ\,\big)$ of $\mathbb{R}^3$ where the law $\,\circ\,$ is defined by
$$ \big(x,\,y,\,z\big)\circ\big(x',\,y',\,z'\big) = \big(x+x',\,z+z'+xy',\,y+y'\big) $$
We denote the left-translation by $\,L_g: p \to g\circ p$.
Considering a tangent vector $\,v=\big(v_1,\,v_2,\,v_3\big)\,$ of $T_xG$, is it easy to compute $\big(dL_g\big)_p$ ?
I mean do we have $$\big(dL_g\big)_p(v) = v_1\big(1,\,0,\,0\big) + v_2\big(0,\,x,\,1\big)+v_3\big(0,\,1,\,0\big) = \big(v_1,\,x\,v_2+v_3,\,v_2\big)\; ?$$
Instead of starting with a tangent vector in $T_xG$, you want to compute the Jacobian of $L_g$ at point $p$. Remember that $T_xG$ is isomorphic to $\mathbb{R}^3$, so you have any vector contained in there. Now
$$ (dL_g)_p \;\; =\;\; \left[ \begin{array}{ccc} \partial_{g_1} (g_1 + p_1) & \partial_{g_2}(g_1 + p_1) & \partial_{g_3}(g_1 + p_1) \\ \partial_{g_1}(g_3 + p_3 + g_1p_2) & \partial_{g_2}(g_3 + p_3 + g_1p_2) & \partial_{g_3}(g_3 + p_3 + g_1p_2) \\ \partial_{g_1}(g_2 + p_2) &\partial_{g_2}(g_2 + p_2) & \partial_{g_3}(g_2 + p_2) \end{array} \right ] \;\; =\;\; \left [ \begin{array}{ccc} 1 & 0 & 0 \\ p_2 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right ]. $$
Then for any vector $v = [v_1 \; v_2 \; v_3 ]^T$ we have
$$ (dL_g)_p(v) \;\; =\;\; \left [ \begin{array}{c} v_1 \\ p_2v_1 + v_3 \\ v_2\\ \end{array}\right ]. $$