I just read the same paper. Initially I made the same mistake as you, which is how I came across your question. It's been more than a year since your post but, in case you are still curious, here is how to work out the derivation. The key is to properly compute $\frac{\partial \mu_n}{\partial x_{ij}}$ for n=2,3, which requires pulling $x_{ij}$ out from the sum:
$$ \begin{align} \frac{\partial \mu_n}{\partial x_{ij}} &= \frac{1}{L} \frac{\partial}{\partial x_{ij}}\Big(\sum_{p,q \neq i,j}(x_{pq} - \mu_1)^n + (x_{ij}-\mu_1)^n\Big)\\ &= \frac{1}{L} \Big(\sum_{p,q \neq i,j}-n(x_{pq} - \mu_1)^{n-1} \frac{\partial}{\partial x_{ij}} \mu_1 + n(x_{ij}-\mu_1)^{n-1}(1-\frac{\partial}{\partial x_{ij}} \mu_1)\Big)\\ &= \frac{1}{L} \Big(\frac{-n}{L}\sum_{p,q} (x_{pq} - \mu_1)^{n-1} + \frac{n}{L}x^{n-1}_{ij} + n(x_{ij}-\mu_1)^{n-1}\frac{(L-1)}{L}\Big)\\ &= \frac{1}{L} \Big(\frac{-n}{L}\sum_{p,q} (x_{pq})^{n-1} + nx^{n-1}_{ij}\Big) \quad \textrm{[use fact that }\mu_1=0]\\ &= \frac{n}{L}(x^{n-1}_{ij} - \mu_{n-1}) \end{align} $$$$ \begin{align} \frac{\partial \mu_n}{\partial x_{ij}} &= \frac{1}{L} \frac{\partial}{\partial x_{ij}}\Big(\sum_{p,q \neq i,j}(x_{pq} - \mu_1)^n + (x_{ij}-\mu_1)^n\Big)\\ &= \frac{1}{L} \Big(\sum_{p,q \neq i,j}-n(x_{pq} - \mu_1)^{n-1} \frac{\partial}{\partial x_{ij}} \mu_1 + n(x_{ij}-\mu_1)^{n-1}(1-\frac{\partial}{\partial x_{ij}} \mu_1)\Big)\\ &= \frac{1}{L} \Big(\frac{-n}{L}\sum_{p,q\neq i,j} (x_{pq} - \mu_1)^{n-1} + n(x_{ij}-\mu_1)^{n-1}\frac{(L-1)}{L}\Big)\\ &= \frac{1}{L} \Big(\frac{-n}{L} \sum_{p,q} x_{pq}^{n-1} + n x^{n-1}_{ij}\Big) \quad \textrm{[use fact that }\mu_1=0]\\ &= \frac{n}{L}(x^{n-1}_{ij} - \mu_{n-1}) \end{align} $$
Subbing the above into:
$$\frac{\partial \eta(x)}{\partial x_{ij}} = \mu^{-3/2}_2\frac{\partial \mu_3}{\partial x_{ij}} - \frac{3}{2}\mu_3\mu^{-5/2}_2\frac{\partial \mu_2}{\partial x_{ij}}$$
Gives:
$$\frac{\partial \eta(x)}{\partial x_{ij}} = \mu^{-3/2}_2\frac{3}{L}(x^2_{ij} - \mu_2) - \frac{3}{2}\mu_3\mu^{-5/2}_2\frac{2}{L}x_{ij}$$
Factoring out $\mu^{-3/2}_2\frac{3}{L}$ gives:
$$\frac{\partial \eta(x)}{\partial x_{ij}} = \mu^{-3/2}_2\frac{3}{L}(x^2_{ij} - \eta(x)\mu^{1/2}_2 x_{ij} - \mu_2)$$
which is the what is reported in the paper.