In the post Differential of the multiplication and inverse maps on a Lie group, second answer, appears the following solution for the push-forward of the inverse map:
$\iota_{*,a}(Y_a) = -(r_{a^{-1}})_* (\ell_{a^{-1}})_* Y_a \tag1$
In the second answer it is explained that applying the chain rule to
$\iota(x) = r_{a^{-1}} \circ \iota \circ l_{a^{-1}}(x) \tag2$
you obtain Eq. (1). I guess that the chain rule has to be applied by derivation respect to some parameter $t$. Doing that I get:
$$\frac{d\iota(x)}{dt} = \frac{dr_{a^{-1}}(\iota(l_a^{-1}(x)))}{d(\iota(l_a^{-1}(x)))}\cdot \frac{d(\iota(l_a^{-1}(x)))}{d(l_{a^{-1}}(x))}\cdot \frac{dl_{a^{-1}}(x)}{dx}\cdot \frac{dx}{dt} = r_{a^{-1}}\frac{d(\iota(l_a^{-1}(x)))}{d(l_{a^{-1}}(x))}l_{a^{-1}}\frac{dx}{dt} \tag3$$
And I don't know how to continue, so, I don't see how Eq. (1) can be obtained by this method.