Your approach is wrong because the left hand side represents the derivative of the composition of two time dependent functions. You completely ignored that fact and started cancelling terms. I have a different solutiontry to this problem in mind.give a simpler example:

$$\frac{d 4x^3}{dx} = kx^2 \Leftrightarrow 12x^2=kx^2 \Leftrightarrow 12=k$$

This is not the same as

$$\frac{d 4x}{dx} = k \Leftrightarrow 4=k$$

Therefore use the chain rule

$$\frac{d(c \cdot V)}{dt} = V\frac{dc}{dt}+\frac{dV}{dt}c=-Ak\Delta c$$

I would now assume a stationary diffusion, hence

$$V\frac{dc}{dt} = 0$$

$$\frac{dV}{dt}c=-Ak\Delta c$$

$$A = \frac{dV}{dr}$$

$$\frac{dV}{dt}c=-\frac{dV}{dr}k\Delta c$$

For this step I hope there are no mathematicians around, I cancel the differentials

$$\frac{dr}{dt}c=-k\Delta c$$

$$\frac{dr}{dt}=-k\frac{\Delta c}{c}$$

I have a different solution to this problem in mind.

$$\frac{d(c \cdot V)}{dt} = V\frac{dc}{dt}+\frac{dV}{dt}c=-Ak\Delta c$$

I would now assume a stationary diffusion, hence

$$V\frac{dc}{dt} = 0$$

$$\frac{dV}{dt}c=-Ak\Delta c$$

$$A = \frac{dV}{dr}$$

$$\frac{dV}{dt}c=-\frac{dV}{dr}k\Delta c$$

For this step I hope there are no mathematicians around, I cancel the differentials

$$\frac{dr}{dt}c=-k\Delta c$$

$$\frac{dr}{dt}=-k\frac{\Delta c}{c}$$