Pseudo Proof for a property of the Del operator

Suppose I want to prove the following identity:

$\nabla. (F \times G)= G.(\nabla \times F) - F.(\nabla \times G)$ for a vector on $\ \mathbb{R}^3$

I know that the "most correct" way to prove this is by invoking the del operator on the vector $\ F \times G$. However, I tried to use the pseudo determinant representation of $\ \nabla \times G$ and $\ \nabla . G$ which I thought would help me to prove these identities faster. However, I realised that some of the time this works, while for other times, this does not work.

Particularly, if you see $\nabla. (F \times G)$ as a determinant of a 3 x 3 matrix as in my working on the LHS, this actually works. However, if we see $\ G.(\nabla \times F)$ as such and apply properties of determinants, it does not seem to work. At least I feel that the step on applying an elementary row operation is really suspicious but I can't point how it is so...

Can anyone enlighten me on why this works sometimes, while it doesnt in other cases? Thanks!

Link to image for my working

For example, usually $P_F\tfrac{\partial ~}{\partial y}R_G \neq R_G\tfrac{\partial ~}{\partial y}P_F$.
So the $R_1\leftrightarrow R_3$ exchange is not going to simply change the sign of the determinant.
What you've done is analogous to claiming$$\tfrac {\mathrm d ~~}{\mathrm d ~x}\big(f(x)g(x)\big)~{= f(x)\big(\tfrac{\mathrm d ~~}{\mathrm d ~x}g(x)\big)+\big(\tfrac{\mathrm d ~~}{\mathrm d ~x}f(x)\big)g(x) \\ = \tfrac{\mathrm d ~~}{\mathrm d ~x}\big(f(x)g(x)\big)+\tfrac{\mathrm d ~}{\mathrm d ~x}\big(f(x)g(x)\big)\\=2\tfrac{\mathrm d ~}{\mathrm d ~x}\big(f(x)g(x)\big)}$$