= Solution
The first coordinate is fixed by the embedded <SU(5)>, while the last five transform in its defining <representation>. Therefore $\mathbf6\downarrow_{SU(5)}=\mathbf1\oplus\mathbf5$, and taking the <dual representation> gives $\overline{\mathbf6}\downarrow_{SU(5)}=\mathbf1\oplus\overline{\mathbf5}$.
For the <symmetric square>, the fixed-fixed component is one <trivial representation>, the fixed-five mixed components give $\mathbf5$, and the symmetric five-five block is $\mathbf{15}$. For the <exterior square>, the fixed-fixed component vanishes, the mixed components give $\mathbf5$, and the antisymmetric five-five block is $\mathbf{10}$. Finally restrict $\mathbf{35}=\mathbf6\otimes\overline{\mathbf6}-\mathbf1$, using $\mathbf5\otimes\overline{\mathbf5}=\mathbf{24}\oplus\mathbf1$.
The five <SU6 tensor branchings to SU5> are thus
$$
\boxed{\begin{aligned}
\mathbf6&\downarrow=\mathbf5\oplus\mathbf1,\\
\overline{\mathbf6}&\downarrow=\overline{\mathbf5}\oplus\mathbf1,\\
\mathbf{21}&\downarrow=\mathbf{15}\oplus\mathbf5\oplus\mathbf1,\\
\mathbf{15}&\downarrow=\mathbf{10}\oplus\mathbf5,\\
\mathbf{35}&\downarrow=\mathbf{24}\oplus\mathbf5\oplus\overline{\mathbf5}\oplus\mathbf1.
\end{aligned}}
$$
The symbols $\mathbf{15}$ on the left and right refer to different <groups>: the left-hand one is the <SU(6)> <exterior square>, whereas the right-hand one is the <SU(5)> <symmetric square>. The <dimension> checks are $6=5+1$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.
Explicitly, in the <Adjoint representation> the lower-right traceless five-by-five block gives $\mathbf{24}$; the lower-left column and upper-right row give $\mathbf5$ and $\overline{\mathbf5}$; and the diagonal direction $\operatorname{diag}(-5,1,1,1,1,1)$ gives the remaining <trivial representation>.
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