Good shot.
The basics are right: $V=(1/3)\pi r^2 h$ and $h=14r/5.$
For part (a) I agree with you:
$$V(h) = \frac{25 \pi h^3}{588} $$
So,
$$\frac{dV(h)}{dt} = \frac{dV(h)}{dh} \frac{dh}{dt} = \frac{25 \pi h^2}{196}\frac{dh}{dt} = -2,$$
and
$$\frac{dh}{dt} = -\frac{392}{25 \pi h^2},$$
which at $h=6$ (feet) is
$$\frac{dh}{dt} = -\frac{98}{225 \pi}$$ feet per second.
For part (b) it seems like you're doing more work than necessary. We are looking for the rate of change in radius at the same height we found the rate of change in height for. But we know that $h = 14r/5$ so
$$\frac{dh}{dt} = \frac{14}{5} \frac{dr}{dt} \to \frac{dr}{dt} = \frac{5}{14} \frac{dh}{dt}.$$
So multiply your answer in (a) by $5/14$ and you're done.
So we know the following:
Rate at which water is entering cone, lets call that $\dfrac{df}{dt}$.
Rate at which the height of the water in the cone is rising, $\dfrac{dh}{dt}$.
We need to find the leak rate, call it $\dfrac{dk}{dt}$.
My hint was that the change in volume of water in the tank, $\dfrac{dv}{dt}$, satisfies
$\dfrac{dv}{dt}=\dfrac{df}{dt}-\dfrac{dk}{dt}$
We have only one of the things we need, but we can find $\dfrac{dv}{dt}$ using our other given (this is the trickiest part). Note that the cone has fixed proportions (the relationship between $r$ and $h$ is a constant fixed ratio). Hence we can rewrite $v=\dfrac{1}{3}\pi r^2h$ entirely in terms of $h$. This uses the dimensions of the tank given in the problem. Hence $r=\dfrac{1}{4}h$. Substituting for $r$, we obtain
$v=\dfrac{1}{3}\pi \left( \dfrac{1}{4}h \right)^2h=\dfrac{1}{48}h^3$
Differentiate implicitly with respect to time
$\dfrac{dv}{dt}=\dfrac{3}{48}h^2\dfrac{dh}{dt}$
But we know what $h$ and $\dfrac{dh}{dt}$ are from the givens in the problem (you found $h$ yourself).
Thus you can find $\dfrac{dv}{dt}$ and use that and then given for $\dfrac{df}{dt}$ to solve for $\dfrac{dk}{dt}$.
Best Answer
Your volume expression is not correct. You have not defined the variable $t$, but it appears to be time (since when?) You need to compute the volume of water as a function of $h$, the height above the tip. Then we are given $h=4, \frac {dh}{dt}=0.2$ You can use $\frac {dV}{dt}=\dfrac {\frac {dh}{dt}}{\frac {dV}{dh}}$ to get $\frac {dV}{dh}$, the rate water is flowing out. Then divide by the area of the top surface of the barrel to get the rate the water is rising.