$\dfrac{dV}{dt}$ is the rate of change of the volume. In particular, $\dfrac{dV}{dt}=1$ cm per second. $\dfrac{dr}{dt}$ is the rate of change of the radius, which is what we want to know.
Now, the surface area of the sphere is actually $4\pi r^2.$ Observe, then, that $$\frac{dV}{dt}=\frac{d}{dt}\left[\frac43\pi r^3\right]=\frac43\pi\frac{d}{dt}\left[r^3\right]=\frac43\pi\cdot 3r^2\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}.$$ Can you take it from there?
Given the multitude of different comments and given the first answer saying that your formula "hold in general", I would like to emphasize: It is NOT true that $V'(r)=S(r)$ for the Volume $V$ and Surface $S$ of general convex bodies scaled by $r$.
What is true, is that locally
$S\approx V/h$,
where locally means that we are considering a small piece of surface, small enough that it is approximately a piece of a plane, and denote by $S$ its area and by $V$ the volume of the body that you get from extending the surface along its normal direction by length $h$ (normal = orthogonal to the surface)
When you want to consider the global case, that is you want to "sum up" the local formulae, the sphere is a special case because when you enlarge the sphere then all small pieces of surface as described above are by the same normal length $h$. For other convex bodies the summing up does not work.
The problem with general convex bodies is that when you enlarge the body some of the small pieces of surface are extended more than others. For example, if you try to apply your formula to rectangles with side lengths $2r$ and $r$ (and center in the origin), you get the încorrect formula $S(r)=V'(r)=(2r^2)'=4r$. The difference to the correct formula $S(r)=6r$ comes from the fact that dividing by $h$, as you do, is "unjust" to the long sides. These long sides get extended by only $h/2$ in their normal direction when $r$ is enlarged to $r+h$.
Best Answer
The surface areas combine to make the constraint:
$$6 a^2 + 4 \pi r^2 = 1$$
where $a$ is the side of the cube and $r$ is the radius of the sphere. Use this to maximize
$$V(a) = a^3 + \frac{4 \pi}{3} r(a)^3$$
where $r=r(a)$ is determined form the surface area constraint. You maximize by taking the derivative of $V$ and setting it to zero:
$$V'(a) = 3 a^2 + 4 \pi r(a)^2 r'(a) = 0$$