Variable liquid flow rate control

Your solution is correct but an easier method (I think) is to use superposition, first we will suppress r ( i.e. we will ignore r ), then get the transfer function, then suppress d then get the other transfer function, then sum the two up to get the final transfer function.

When we suppress r will then get

$$ \frac{Y}{d} = \frac{P}{1 + PC} $$

and when we suppress d we will get $$ \frac{V}{r} = \frac{C}{1 + PC} $$

but Y = VP meaning

$$ \frac{Y}{r} = \frac{PC}{1 + PC} $$

We the sum up the two values of Y to get the final response of

$$ Y = \frac{P}{1 + PC}d + \frac{PC}{1 + PC}r $$

There is no transfer function from d to y because y depends on both d and r, you can't evaluate the value y without knowing both d and r values.

From filter theory we know that a second-order transfer function is overdamped for a pole quality factor Qp<0.5 (two real poles on the neg. real axis); it is underdamped for Qp>0.5 (a conjugate-complex pole pair in the left half of the s-plane).

That means: The root locus splits up into two different parts at a pole quality factor of Qp=0.5. This is - by definition - a critically damped system. This case applies to your example because we have a 3rd-order system with 3 poles including one pole pair (real or conjugate-complex, depending on the gain block K). As a consequence, the system is critically damped when the gain reaches a value K=Kc corresponding to a value Qp=0.5, which is identical to the "breakaway point".

As a simple and intuitive example: Two decoupled equal RC sections provide a quality factor of Qp=0.5 and exhibit a "critically damped" response.