A paper I am reading uses an identification of $GL(2,\mathbb{R})$ representations:
\begin{align} \Lambda^2(\text{Sym}^3\mathbb{R}^2) = (\text{Sym}^4\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2 ) \oplus (\Lambda^2\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2) \end{align}
which wasn't obvious to me. I've tried solving this the following way: First write
\begin{align} \otimes^2 (\text{Sym}^3\mathbb{R}^2) = \Lambda^2(\text{Sym}^3\mathbb{R}^2) \oplus \text{Sym}^2(\text{Sym}^3\mathbb{R}^2) \end{align}
Now I -think- there is an isomorphism of representations
\begin{align} \phi:\otimes^2 (\text{Sym}^3\mathbb{R}^2) \cong \text{Sym}^6\mathbb{R}^2 \oplus(\text{Sym}^4\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2 ) \oplus (\text{Sym}^2\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2) \oplus (\Lambda^2\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2) \end{align} where all the terms in the decomposition are irreducible representations. Then given this isomorphism, the image of the 6-dimensional $GL(2,\mathbb{R})$ invariant subspace $\Lambda^2(\text{Sym}^3\mathbb{R}^2))$ must be the invariant subspace $ (\text{Sym}^4\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2 ) \oplus (\Lambda^2\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2 \otimes \Lambda^2\mathbb{R}^2)$ by a count on dimensions.
My problem is I can't find the isomorphism $\phi$. Given a tensor $T \in \otimes^2 (\text{Sym}^3\mathbb{R}^2)$ with components $T^{abcdef}$ with $T^{(abc)(def)} = T^{abcdef}$. I have tried the obvious map, namely: Total symmetrisation in the first component of the map $\phi$, symmetrisation $(abde)$ and antisymmetrisation $[cf]$ in the second component, symmetrisation $(ad)$ and antisymmetrisation $[be]$ $[cf]$ in the third component. Then in the last component $[ad]$ $[be]$ $[cf]$. But I can't see why this gives an isomorphism, even when I try to enforce the required symmetries on $T$ such that $T$ is in the kernel of the map I just described.