You want $\dfrac{dh}{dt}$; by the chain rule this is $\dfrac{dh}{dv}\dfrac{dv}{dt}$.
You have $h=\dfrac{v}{\pi r^2}=\dfrac1{\pi r^2}v$, where $\dfrac1{\pi r^2}$ is a constant, so $\dfrac{dh}{dv}=\dfrac1{\pi r^2}$; you don't need the quotient rule for this differentiation. Finally, you have $\dfrac{dv}{dt}=3$, so $$\frac{dh}{dt}=\frac{dh}{dv}\frac{dv}{dt}=\frac3{\pi r^2}\text{ m/min}\;.$$ Since $r=5$ m, the actual rate is $\dfrac3{25\pi}$ m/min.
In a problem like this it's a good idea to use the $\dfrac{dv}{dt}$ notation instead of the $v'$ notation, because you're taking derivatives with respect to more than one variable, and you need to keep track of what that variable is in each calculation.
Let $h(t)$ be the height of the water at time $t$, and let $r(t)$ be the radius of the surface of the water at time $t$; from similar triangles we know that
$$\frac{h(t)}{r(t)}=\frac{14}{2.75}=\frac{56}{11}\;.$$
We could solve this for either $r(t)$ or $h(t)$ in terms of the other, but notice that we’re told $h'(t)$ at a particular moment, and we’re not told anything about an specific value of $r(t)$. This suggests that we’d be better off working in terms of $h(t)$, so we’ll solve for $r(t)$ and get $$r(t)=\frac{11}{56}h(t)\;.$$
At time $t$ the volume $V(t)$ of water in the tank is the volume of a right circular cone with height $h(t)$ and base radius $r(t)$, which is given by
$$V(t)=\frac13\pi r(t)^2h(t)=\frac{\pi}3\left(\frac{11}{56}h(t)\right)^2h(t)=\frac{121\pi}{9408}h(t)^3\;.$$
Then
$$V'(t)=\frac{121\pi}{3136}h(t)^2h'(t)\;.$$
We’re told that $h'(t)=0.24$ when $h(t)=3$; if we call that moment time $t_0$, we have
$$V'(t_0)=\frac{121\pi}{3136}\cdot3^2\cdot0.24=\frac{3267\pi}{39200}\text{ m}^3/\text{min}\;.$$
Now let $v$ be the rate in cubic metres per minute at which water is being pumped into the tank. Taking into account both the inflow and the leakage, we know that at all times
$$V'(t)=v-0.0068\text{ m}^3/\text{min}\;.$$
In particular, at time $t_0$ we have
$$v-0.0068=\frac{3267\pi}{39200}\;,$$
which is completely straightforward to solve for $v$.
Best Answer
Each quantity is a function of time except for the radius $r$ because it does not change with time (well, technically speaking, you still could think of it as a function of time, but it would be a constant function then: $r(t)=5\ m$). All those quantities are related by this expression:
$$V(t)=\pi r^2 h(t)$$
This equality states that at any given time, the volume of the water in the tank as a function of time equals the height of the water in the tank as a function of time multiplied by $\pi r^2$. You know the rate at which the volume of the water in the tank is increasing and you know what $r$ is. You also know that $V(t)$ is equivalent to $\pi r^2 h(t)$. If they are equivalent, their derivatives must also be equivalent:
$$V'(t)=[\pi r^2 h(t)]'=\pi r^2 h'(t)$$
Now, just solve for $h'(t)$ which tells you exactly what the problem is asking you to find—how fast the height of the water in the tank increases:
$$h'(t)=\frac{V'(t)}{\pi r^2 }=\frac{3}{25\pi}\ m/min.$$