Derivative of $(uA+C)^{-1}\mathbf{b}$ w.r.t. $u\in\mathbb{R}$

Define $\,M=(C+uA)\,$ then the given equation becomes $\,Mx=b$

Differentiate the equation (with respect to $u)\,$ then solve for $\dot x=\left(\frac{dx}{du}\right)$ $$\eqalign{ \dot Mx + M\dot x = \dot b \\ Ax + M\dot x = 0 \\ \dot x = -M^{-1}Ax \\ }$$ This is indeed the implicit differentiation technique that you remembered.

Hint Since $b$ does not depend on $u$, $$\frac{d}{du}[(u A + C)^{-1} {\bf b}] = \frac{d}{du}[(u A + C)^{-1}] {\bf b} ,$$ and so it suffices to know how to compute the derivative $\frac{d}{du} [P(u)^{-1}]$ inverse of a matrix function $$P : \Bbb R \to M_n (\Bbb R)$$ (wherever that inverse is defined).

We we can find $\frac{d}{du}(P(u)^{-1})$ in terms of $P$ and $\frac{d P}{dt}$ by differentiating both sides of $P(u) P(u)^{-1} = I$ and isolating $\frac{d}{du}[P(u)^{-1}]$.

Suppressing the argument $u$, we have $$\frac{dP}{du} P^{-1} + P \frac{d}{du} (P^{-1}) , $$ so $$\frac{d}{du} (P^{-1}) = - P^{-1} \frac{dP}{du} P^{-1} .$$