To prove an identity $x=y$ of vectors in $\Bbb R^3$, it suffices to prove that
$x\cdot c=y\cdot c$ for all $c\in\Bbb R^3$. So for $(2)$, I'll prove that
$$(b \times (Ma) - a \times (Mb))\cdot c =
(M^T (a \times b) - (\operatorname{tr}{M})(a \times b))\cdot c . \tag{2'}$$
for all vectors $a$, $b$ and $c$.
Write
$$[a,b,c]=(a\times b)\cdot c$$
for the scalar triple product. This is a trilinear map, and is alternating:
$$[b,b,c]=[a,b,b]=0$$
for all $a$, $b$, $c$. There is only one alternating trilinear map up to scalar
multiplication; any alternating trilinear map is a constant multiple
of the scalar triple product.
Let $M$ be a matrix, and define
$$\phi(a,b,c)=[Ma,b,c]+[a,Mb,c]+[a,b,Mc].$$
It's easy to see that $\phi$ is alternating trilinear, so
$$\phi(a,b,c)=t[a,b,c]$$
where $t$ depends only on $M$. Taking $a$, $b$, $c$ to be the standard unit
vectors gives $t=\text{Tr}(M)$. Therefore
$$[Ma,b,c]+[a,Mb,c]+[a,b,Mc]=\text{Tr}(M)[a,b,c].\tag{3}$$
Observe that
$$[Ma,b,c]=-(b\times Ma)\cdot c,$$
$$[a,Mb,c]=(a\times Mb)\cdot c,$$
$$[a,b,Mc]=(a\times b)\cdot Nc=(M^T(a\times b))\cdot c$$
(using the identity $x\cdot My=M^Tx\cdot y$)
and
$$\text{Tr}(M)[a,b,c]=(\text{Tr}(M)(a\times b))\cdot c.$$
So $(3)$ becomes
$$(-b\times Ma+a\times Mb+M^T(a\times b))\cdot c=\text{Tr}(M)(a\times b)\cdot c$$
which is equivalent to $(2')$.
Best Answer
The formula you derived reads $$u \times (\nabla \times v) = \nabla_v(u \cdot v) - (u \cdot \nabla)v$$ where the notation $\nabla_v$ is called Feynman notation and should indicate that the derivative is applied only to $v$ and not to $u$.