For practice I wanted to define Lie bracet in terms of coordinates and show that this definition is independet of the choice of coordinates. So I started with the definition $$[X,Y]:=X^i\partial_i(Y^j)\partial_j-Y^j\partial_j(X^i)\partial_i,$$ where $\partial_i=\frac{\partial}{\partial x^i}$ and $X=X^i\partial_i,Y=Y^j\partial_j$ taking into account Einstein notation.
Now we take new coordinates $\bar\partial_k=\frac{\partial}{\partial y^k}.$ So we have that $$X^i=J^i_k\bar X^k,\hspace{10pt} Y^j=J^j_l\bar Y^l \hspace{10pt}\text{and} \hspace{10pt}\partial_i=P_i^s\bar\partial_s$$ where $J^i_k=\frac{\partial x^i}{\partial y^j}$ and $P^s_i=\frac{\partial y^s}{\partial x^i}.$ I put above to my bracket and get $$\bar X^k J^i_kP^s_i\bar\partial_s(\bar Y^lJ_l^j)P^r_j\bar\partial_r-\bar Y^l J^j_lP^r_j\bar\partial_r(\bar X^kJ_k^i)P^s_i\bar\partial_s.$$ $J^j_lP^r_j=\delta_l^r$ hence $$\bar X^k\bar\partial_k(\bar Y^lJ_l^j)P^r_j\bar\partial_r-\bar Y^l \bar\partial_l(\bar X^kJ_k^i)P^s_i\bar\partial_s.$$ Using Leibniz rule to $\bar\partial_k$ and $\bar\partial_l$ I get $$\bar X^k\bar\partial_k(\bar Y^l)J_l^jP^r_j\bar\partial_r+\bar X^k\bar Y^l\bar\partial_k(J_l^j)P^r_j\bar\partial_r-\bar Y^l \bar X^k\bar\partial_l(J_k^i)P^s_i\bar\partial_s-\bar Y^l \bar\partial_l(\bar X^k)J_k^iP^s_i\bar\partial_s.$$ Using yet $J^j_lP^r_j=\delta_l^r$ twice I get $$\color{blue}{\bar X^k\bar\partial_k(\bar Y^l)\bar\partial_l-\bar Y^l \bar\partial_l(\bar X^k)\bar\partial_k}+\bar X^k\bar Y^l\bar\partial_k( J_l^j)P^r_j\bar\partial_r-\bar Y^l \bar X^k\bar\partial_l(J_k^i)P^s_i\bar\partial_s.$$ Now the blue part is what I want. The rest should vanish.
Question How to make the rest vanish?