I want to show that $$\tau(n)=\frac{n}{12}(5\sigma_3(n)+7\sigma_5(n))-70\sum_{m=1}^{n-1}(5m-2n)\sigma_3(m)\sigma_5(n-m)$$ is true for all $n\in \mathbb{N}$. Previously I showed that $-147G_6^2+\frac{200}{3}G_8G_4=\Delta$. So somehow I have to derive the formula from this. I used the cauchy product formula to show $$-147G_6^2=-147\sum_{n=0}^\infty\sum_{k=0}^n\sigma_5(k)\sigma_5(n-k)$$ and $$\frac{200}{3}G_8G_4=\frac{200}{3}\sum_{n=0}^\infty \sum_{k=0}^n\sigma_7(k)\sigma_3(n-k)$$
EDIT: I tried another approach now but still cannot find the right solution.
I showed that $$(-35)\frac{1}{2\pi}(6G_4'G_6-4G_4G_6')=\Delta$$
and got the following expansions $\frac{1}{2\pi} G_4'=\sum_{n=1}^\infty n\sigma_3(n)q^n$ and $\frac{1}{2\pi} G_6'=\sum_{n=1}^\infty n\sigma_5(n)q^n$. Then I used the Cauchy-product and got $$\frac{1}{2\pi} 6G_4'G_6=6\sum_{n=1}^\infty \left(\sum_{k=0}^n k\sigma_3(k)\sigma_5(n-k)\right)$$
$$\frac{1}{2\pi} 4G_4G_6'=4\sum_{n=1}^\infty \left(\sum_{k=0}^n \sigma_3(k)\sigma_5(n-k)(n-k)\right)$$ And after multiplying with the $-35$ I get to
$$\tau(n)=140 \sum_{k=0}^n \sigma_3(k)\sigma_5(n-k)(n-k)-210\sum_{k=0}^n \sigma_3(k)\sigma_5(n-k)n$$ which simplifies to
$$\tau(n)=-70\sum_{k=0}^n n\sigma_3(k)\sigma_5(n-k)-140\sum_{k=0}^n k\sigma_3(k)\sigma_5(n-k)$$
And from here on I do not know how to proceed.especially I can't see how to get $(5k-2n)$ and the first summand. And it seems that this is already wrong since according to this formula is $\tau(1)\neq1$