Why does potential energy time-derivative depend only on the mass flow?

For your problem, $$dot{W_P}=dot{m}[(g z)_{top}-(gz)_{bottom}]$$where z is the elevation and $dot{m}$ is the mass flow rate being pumped. $dot{W_P}$ is the rate at which the pump does shaft work on the fluid.

First of all, about power. It is the rate of doing work, and more complicated than defining it as a rate of total energy.

Now, your working was not off. All is technically correct but $h$ does not actually change since the water behaves as a stationary column of water. If $frac{dh}{dt}neq0$ then this would mean that the height difference is changing.

Remember that $dh$ is similar to $Delta h$ in that it represents a change in height of the water (in this case the height difference). We know that this is a constant at $15 space m$, so there is no change over time. Thus $frac{dh}{dt}=0$.

We can plug this in your last step: $$P=mcdot g cdot frac {dh}{dt} + g cdot h cdot ρcdot frac {dV}{dt}$$ $$=mcdot g cdot 0 + g cdot h cdot ρcdot frac {dV}{dt}$$ $$therefore space =g cdot h cdot ρcdot frac {dV}{dt}$$

Once realized that $h$ is a constant, the chain rule is no longer necessary and can get to this result from an earlier step: $$frac{d(mgh)}{dt}$$ Factor out constants: $$=gcdot h cdot frac{dm}{dt}$$ $$therefore space =g cdot h cdot ρcdot frac {dV}{dt}$$