I think I am close with this proof - I just need the final step. Let $R(\mathbf{n}, \theta) \in SO(3) $ be the 3D rotation matrix of angle $\theta$ around axis $\mathbf{n}$. Then show that $$\frac{d}{d\theta} R(\mathbf{n}, \theta) = S(\hat{\mathbf{n}})R(\mathbf{n}, \theta)$$ where $\hat{\mathbf{n}}$ is the unit vector lying on the axis $\mathbf{n}$, and $S(\mathbf{x})$ is the skew-symmetric matrix $$ S(\mathbf{x}) = \begin{pmatrix} 0 & -x_3 & x_2 \\ x_3 & 0 & -x_1\\ -x_2 & x_1 & 0 \end{pmatrix}$$ We have $$R(\mathbf{n}, \theta)R(\mathbf{n}, \theta)^T = \mathbf{I}$$ so $$\frac{d}{d\theta}(R(\mathbf{n}, \theta)R(\mathbf{n}, \theta)^T) = \frac{d}{d\theta}\mathbf{I}$$
$$(\frac{d}{d\theta}R(\mathbf{n}, \theta))R(\mathbf{n}, \theta)^T + R(\mathbf{n}, \theta)(\frac{d}{d\theta}R(\mathbf{n}, \theta))^T = \mathbf{0}$$
$$(\frac{d}{d\theta}R(\mathbf{n}, \theta))R(\mathbf{n}, \theta)^T + ((\frac{d}{d\theta}R(\mathbf{n}, \theta))R(\mathbf{n}, \theta)^T)^T = \mathbf{0}$$
which implies $$S + S^T = \mathbf{0}$$ for $$S = (\frac{d}{d\theta}R(\mathbf{n}, \theta))R(\mathbf{n}, \theta)^T$$
So $S$ is a skew-symmetric matrix, and $$\frac{d}{d\theta}R(\mathbf{n}, \theta) = SR(\mathbf{n}, \theta)$$
$\hat{\mathbf{n}}$ is an eigenvector of $R(\mathbf{n}, \theta)$, so multiplying both sides by $\hat{\mathbf{n}}$, we have: $$\frac{d}{d\theta}R(\mathbf{n}, \theta)\hat{\mathbf{n}} = SR(\mathbf{n}, \theta)\hat{\mathbf{n}}$$ $$\frac{d}{d\theta}(R(\mathbf{n}, \theta)\hat{\mathbf{n}}) = S(R(\mathbf{n}, \theta)\hat{\mathbf{n}})$$ $$\frac{d}{d\theta}\hat{\mathbf{n}} = S\hat{\mathbf{n}}$$ $$\mathbf{0} = S\hat{\mathbf{n}}$$
Solving for S, we find: $$ S = \begin{pmatrix} 0 & -a\hat{\mathbf{n}}_3 & a\hat{\mathbf{n}}_2 \\ a\hat{\mathbf{n}}_3 & 0 & -a\hat{\mathbf{n}}_1\\ -a\hat{\mathbf{n}}_2 & a\hat{\mathbf{n}}_1 & 0 \end{pmatrix} = aS(\hat{\mathbf{n}})$$ for all real $a$, and so we have: $$\frac{d}{d\theta} R(\mathbf{n}, \theta) = aS(\hat{\mathbf{n}})R(\mathbf{n}, \theta)$$
So it seems that all that there is left to do is show that $a = 1$, but I can't work out how to do this. Taking the determinant and trace seem like algebraic dead ends, because $det(S) = tr(S)= 0$. Any ideas?
Let $N=S(n)$ and recall that $\,\,N^3=-N$.
Using the Rodrigues formula, the proof is simple. $$\eqalign{ R &= I + N\sin\theta + N^2(1-\cos\theta) \cr NR &= N + N^2\sin\theta + N^3(1-\cos\theta) = N^2\sin\theta + N\cos\theta \cr \frac{d}{d\theta}R &= \big(N\cos\theta + N^2\sin\theta\big) = NR \cr }$$