Let primes denote total derivatives with respect to $t$ and subscripts denote partial derivatives, so that $t_\lambda=\lambda_t=0$ but $\lambda'=\frac{d\lambda}{dt}$ is "irreducible" (and not necessarily zero), and assume that all derivatives below exist. Given the equation $ c'=\frac{d}{dt}c(t,\lambda)=f(c,\lambda) $ for the total derivative of $c$ with respect to $t$, note that (a priori) $ dc =\frac{\partial c}{\partial t}dt +\frac{\partial c}{\partial \lambda}d\lambda \quad\implies\quad c'=c_t+c_\lambda\lambda' $ which in our case (a posteriori), i.e. from $f=c'$, implies that $ \frac{\partial c}{\partial \lambda} =c_\lambda=\frac{c'-c_t}{\lambda'} =\frac{f-c_t}{\lambda'}. $ Although $c_t$ and $c_\lambda$ are, like $c(t,\lambda)$, functions of the independent variables $t$ and $\lambda$, they nevertheless also satisfy a constraint with respect to the the total derivative. This is true a priori. However, in our case, we are given that $c$ satisfies an ordinary differential equation with respect to time, i.e. the total derivative of $c$ with respect to "time" $t$ is itself a function $f$ of $c$ and $\lambda$. This is specific to our problem (a posteriori), and introduces extra paths (from $c'$ to $c$ and $\lambda$), but not cycles, in the directed graph or DAG of variable dependencies depicted below. These extra dependency paths are shown in green, while the original depenencies of $c(t,\lambda)$ are red, and all the rest, which follow from a priori principles (like the total derivative a.k.a. chain rule formula are black).

Now from the (a priori) definition of total derivative, $ \frac{d}{dt}f(c,\lambda) =f_cc'+f_\lambda\lambda' =ff_c+\lambda'f_\lambda $ and $ (c_t)'=c_{tt}+\lambda'c_{t\lambda} $ so that (using the quotient rule) $ \frac{d}{dt} \left( \frac{\partial c}{\partial \lambda} \right) = (c_\lambda)' = \left( \frac{f-c_t}{\lambda'} \right)' = \frac{ \left(f'-(c_t)'\right)\lambda' - \left(f-c_t\right)\lambda'' }{(\lambda')^2} $ $ =\frac{ \left(ff_c+\lambda'f_\lambda\right) -\left(c_{tt}+\lambda'c_{t\lambda}\right) -\lambda''c_\lambda }{\lambda'} $ $ =f_\lambda-c_{t\lambda} +\frac{ff_c-c_{tt}-\lambda''c_\lambda}{\lambda'} $ Note that we did not need or use the equations $(c_\lambda)'=c_{\lambda t}+\lambda'c_{\lambda\lambda}$ $(c')_t=c_{tt}+\lambda'c_{\lambda t}$ but by the latter, we can see that $(c_t)'=(c')_t$ $\iff$ $c_{\lambda t}=c_{t\lambda}$ $\iff$ $c(t,\lambda)\in C^2$.
Also, note that
- the answer would just be $(c_\lambda)'=c_{\lambda t}$, by the former equation above, if $\lambda$ were a model parameter so that $\lambda'=0$;
- the answer would be quite different if $f$ were the partial, rather than the total, derivative of $c$, i.e. if we were given $\frac{\partial}{\partial t}c(t,\lambda)=f(c,\lambda)$ in the original problem.