Need help with Applications of Differentiation Problem Question
 
My Working
Following the hint + some help I got from tutorial below is what I got ... but I believe I did something wrong ... its not the answer below the question yet $0.955\text{ }m\text{ }min$ (I believe its $0.955 m/min$)

UPDATE
In general how do I start with such problems without any hints? I suppose I must somehow formulate the equation like:
$$\frac{dX}{dt} = \frac{dX}{dY} \cdot \frac{dY}{dt}$$
So that the $dY$ cancels. But suppose the following question ... 

I did: 
$$\frac{dV}{dt} = 5$$
$$Find \frac{dA}{dt} \text{ when } V = 216 cm^3$$
$$\frac{dV}{dt} = \frac{dA}{dt} \cdot \frac{dV}{dA}$$
So I tried expressing $V$ in terms of $A$: But I still ended up with an $l$
$$V = l^3, A = 6l^2$$
$$V = \frac{1}{6} A l$$
 A: There are a couple of mistakes, at least one of which is a minor slip.  I will use the notation of the post.
The equation
$$V=\frac{1}{3}\pi h^3 \tan^2\theta$$
for $V$ as a function of $h$ is correct.  Since $\tan\theta=\frac{1}{3}$, it would be cleaner to write $V=\frac{1}{27}\pi h^3$.  But we will continue using $\tan^2\theta$.
Now we differentiate both sides with respect to $t$. Here there are two errors in the post. By the Chain Rule, we have
$$\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}=(\pi h^2\tan^2\theta)\frac{dh}{dt}.$$
There was a little slip here, you wrote $2h^2$ for the derivative of $h^3$ with respect to $h$, and it should be $3h^2$. In addition, the necessary $\frac{dh}{dt}$ part, though mentioned correctly once, later appears on the "wrong" side of the equation for $\frac{dV}{dt}$.  The Chain Rule is usually an essential tool in related rates problems, and needs to be handled carefully.
Now that we have a general relationship between $\frac{dV}{dt}$ and $\frac{dh}{dt}$, "freeze" things at the instant $t$ when $r=2$. When $r=2$, we have $h=6$. And while the cone is filling up, $\frac{dV}{dt}=12$. So at the instant when $r=2$, we have
$$12=(\pi)(6^2)(\tan^2\theta)\frac{dh}{dt}.$$
You did some unnecessary work at the corresponding point of your calculation, but it did not result in an error. You found $\theta$ using the calculator, and then $\tan^2\theta$. That is not needed, since you already know that $\tan\theta=\frac{1}{3}$. Whatever approach we take, we should get
$$12=(\pi)(6^2)(1/3)^2\frac{dh}{dt} =4\pi\frac{dh}{dt}.$$
Solve for $\frac{dh}{dt}$. We find that, at the instant when $r=2$,
$$\frac{dh}{dt}=\frac{12}{4\pi}=\frac{3}{\pi}.$$
Numerically, the answer is about $0.95493$.
Added: We look briefly at the "cube" problem that was added to the post.  We have $\frac{dV}{dt}=5$, and want $\frac{dA}{dt}$.  I think the natural thing to do is to let $s=s(t)$ be the side at time $t$. (In the post it is called $l$, fine too, but looks too much like $1$!)
Everybody knows that
$$V=s^3 \text{ and } A=6s^2$.$$
Don't think, differentiate with respect to time. 
$$\frac{dV}{dt}=\frac{dV}{ds}\frac{ds}{dt}=3s^2\frac{ds}{dt} \text{ and } \frac{dA}{dt}=12s\frac{ds}{dt}.$$ 
We have $\frac{dA}{dt}=5$. When $V=216$, $s=6$. From the expression for $\frac{dV}{dt}$ above, we find that $\frac{ds}{dt}=5/108$ when $s=6$. Now we can use the expression for $\frac{dA}{dt}$ to conclude that $\frac{dA}{dt}=(12)(6)(5/108)$. Simplify if desired.
Or else find a direct relationship between $V$ and $h$. We have $V=r^3$, and $A=6r^2$, so $r^2=A/6$. Since $r^3=V$, we have $r^6=V^2=(A/6)^3$, So $216V^2=A^3$. Differentiate with respect to $t$. We get
$$(216)(2V)\frac{dV}{dt}=3A^2\frac{dA}{dt}.$$
When $V=216$, $A=216$. Now we can use the above equation to find $\frac{dA}{dt}$ when $V=216$.
I prefer the first way the problem was done, but the second fits better into the "find a relationship, then differentiate" pattern. 
Comment: The detailed description of what you did was very helpful in locating the problems.  It would be very nice if everybody was this thorough in showing work! 
