For rotation about the $e_3$, or z-axis, let
$$\eqalign{
B &= \omega^\times\,t \cr
b &= \omega_3\,t \cr
G_3 &= B/b \cr
}$$
$G_3$ is a skew-symmetric (0-1) matrix, whose square is
$$\eqalign{
G_3^2 &= e_3e_3^T-I \cr\cr
}$$
Evaluating the exponential formula
$$\eqalign{
R = e^B &= \frac{\sin b}{b}\,B &+\,I + \frac{1-\cos b}{b^2}\,B^2 \cr
&= (\sin b)\,G &+\,I + (1-\cos b)\,G^2 \cr
&= (\sin b)\,G &+\,(\cos b)\,I + (1-\cos b)\,e_3e_3^T \cr
\cr}$$
The first 2 terms yield the familiar $(2\times 2)$ cosine-sine rotation matrix.
The third term sets the lower right element by subtracting the $\cos(b)$ factor coming from the $(3\times 3)$ identity matrix in the second term, and replacing it with a value of 1.
We start, as usual, with the equality $Av=\lambda v$ where $v^Tv=1$ and $A$ is a $C^1$ function. It is absolutely necessary that the considered eigenvalue $\lambda$ is simple - then $\lambda,v$ are $C^1$ function- otherwise, $v$ may be non-continuous.
Proposition. Under the above hypothesis, $\lambda',v'$ are functions of $A,A',\lambda,v$. More precisely,
$\lambda'=v^TA'v,v'=w-(v^Tw)v$ where $w\in(A-\lambda I)^{-1}((v^TA'v)v-A'v)$.
Proof. We obtain $A'v+Av'=\lambda'v+\lambda v',v'^Tv=v^Tv'=0,v^TA=\lambda v^T,v^TAv'=0$.
Moreover, $v^TA'+v'^TA=\lambda'v^T+\lambda v'^T$ implies $v^TA'v=\lambda'$.
$A'v+Av'=(v^TA'v)v+\lambda v'$ implies $(A-\lambda I)v'=(v^TA'v)v-A'v$ and, finally, $v'=w+kv$, where $w\in (A-\lambda I)^{-1}((v^TA'v)v-A'v)$. It remains to calculate $k$; $v^Tv'=v^Tw+kv^Tv$ implies $k=-v^Tw$ and we are done.
Best Answer
You have a function from $\mathbb R^3$ to $M_{33}$ (the set of $3 \times 3$ matrices. Its three partial derivatives are $$ \frac{\partial f}{\partial \omega_1} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix} \\ \frac{\partial f}{\partial \omega_2} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{bmatrix} \\ \frac{\partial f}{\partial \omega_3} = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} . $$
I suppose that I could treat $M_{33}$ as $\mathbb R^9$, and write out a $9 \times 3$ matrix, but would that really be any better?