Vector fields such that their Lie derivatives differ

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I'm trying to find some vectors fields $X$, $Y$, $Z$ on $\mathbb R^2$ such that the Lie derivatives $L_XZ\ne L_YZ$ at the origin, and $X=Y=\frac{\partial}{\partial x}$ on the x-axis.

Now, we have that $X(x,0)=Y(x,0)=\frac{\partial}{\partial x}$. By definition, the Lie derivatives at a point $p\in\mathbb R^2$ are defined as $$(L_XZ)_p=\lim_{t\rightarrow 0}\frac{Z_p-\phi_{t^*}Z_{\phi_t^{-1}(p)}}{t}=-\frac{d}{dt}|_{t=0}(\phi_{t^*}Z_{\phi_t^{-1}(p)})$$ $$(L_XZ)_p=\lim_{t\rightarrow 0}\frac{Z_p-\psi_{t^*}Z_{\psi_t^{-1}(p)}}{t}=-\frac{d}{dt}|_{t=0}(\psi_{t^*}Z_{\psi_t^{-1}(p)})$$ with $\phi$ and $\psi$ being local flows on $X$ and $Y$, respectively. In our case $p=(0,0)$. My idea would be to consider $$X(x,y)=\begin{cases} \dot{x}=1 \\ \dot{y}=y \end{cases}$$ $$Y(x,y)=\begin{cases} \dot{x}=1 \\ \dot{y}=2y \end{cases}$$ Then, we have a closed-form solution $\phi(t)=(x_0+t,y_0e^{t})$ and $\psi(t)=(x_0+t,y_0e^{2t})$, but I can't figure out which $Z$ to pick up.