In this paper in section 4 they define
$$ \bf{h_1} = \bf{x_i} \bf{A}\mbox{; } h_2=h_1C\mbox{; } h_3 = h_2 D\mbox{; } y_i = h_3 C^{-1} $$
where $\bf{A} = diag(\bf{a})$ and $\bf{D} = diag(\bf{d})$. Cosine transform is defined such that $\bf{C}^{-1} = \bf{C}'$.
Then they apply the chain rule for derivatives and write:
$$ \frac{\partial L}{\partial \bf{d}} = \frac{\partial \bf{y_i}}{\partial \bf{d}} \frac{\partial L}{\partial \bf{y_i}} = \frac{\partial \bf{h_2} \bf{D}}{\partial \bf{d}}\frac{\partial \bf{h_3}\bf{C}^{-1}}{\partial \bf{h_3}}\frac{\partial L}{\partial \bf{y_i}} = diag(\bf{h_2})\bf{C} \frac{\partial L}{\partial \bf{y_i}} $$
and also
$$ \frac{\partial L}{\partial \bf{a}} = \frac{\partial \bf{x_i} \bf{A}}{\partial \bf{a}}\frac{\partial \bf{h_1C}}{\partial \bf{h_1}}\frac{\partial\bf{ h_2 D}}{\partial \bf{h_2}}\frac{\partial L}{\partial \bf{h_3}} = \bf{x_i} \odot \bf{C}^{-1}\bf{d} \odot\bf{C}\frac{\partial L}{\partial \bf{y_i}} $$
What I don't understand is how $\frac{\partial \bf{h_3}\bf{C}^{-1}}{\partial \bf{h_3}} = \bf{C}$ and $\frac{\partial \bf{h_1C}}{\partial \bf{h_1}} = \bf{C}^{-1}$.
It' s a problem of convention, they use denominator convention, therefore the result is the transpose of what I'd have said. Being their definition of discrete cosine transform orthogonal, $C^{-1} = C^T$ and $(C^T)^T = C$.