pedja's answer does seem to be expressed in a somewhat complicated way.
Let $A$ be the area in square centimeters. Let $s$ be the length of the side in centimeters. Let $t$ be time in seconds.
Then we are given $\dfrac{ds}{dt} = 6$.
We recall that $A = s^2$.
We want $\dfrac{dA}{dt}$ when $A=16$.
$$
\frac{dA}{dt} = \frac{d}{dt} s^2 = 2s \frac{ds}{dt}.
$$
When $A=16$ then $s=4$ and $ds/dt = 6$. So
$$
2s\frac{ds}{dt} = 2\cdot4\cdot 6.
$$
Disk problem: This one is more mechanical, so let's do it first. Let $r=r(t)$ be the radius of one of its faces. Note that $r$ is changing. We are told how fast $r$ is changing. In symbols, what we are told amounts to
$$\frac{dr}{dt}=0.02.$$
We are asked how fast the area of a face (side, like a coin) is changing. Let $A=A(t)$ be the area of the face. We want to find out about $\dfrac{dA}{dt}$.
We are given the rate of change of something, and want to find the rate of change of something else. We therefore need a link between the two quantities. In our case, the link is through the familiar formula
$$A=\pi r^2.$$
Differentiate both sides with respect to $t$. To differentiate $r^2$, use the Chain Rule. We get
$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}.$$
At the instant when $r=8.1$, we know everything on the right-hand side. At that instant, area is changing at the rate of $(2\pi)(8.1)(0.02)$ (square inches per second).
Plane problem: First step: Draw a diagram. Let $O$ be the position of the observer, and let $A$ be the point $1$ mile above the observer. Let $P$ be the position of the plane. The position $P$ is changing. Note that $\triangle OAP$ is right-angled.
Let $x=x(t)$ be the distance $OP$. This distance is changing. We know at what rate $x$ is changing: it is the speed of the plane. Do we know that
$$\frac{dx}{dt}=400.$$
We are asked how fast the distance of the plane from the observer is changing. So we are asked how fast $OP$ is changing. Let $OP=z$. The $z=z(t)$ is a function of $t$. We want to find out about
$$\frac{dz}{dt}.$$
We need a link between $x$ and $z$. In this case, that is provided by the Pythagorean Theorem. Since $OA=1$, we have
$$z^2=1^2+x^2.$$
Differentiate immediately* with respect to $t$, using the Chain Rule. Now you should be able to finish quickly. We get
$$2\frac{dz}{dt}=2x\frac{dx}{dt},\quad\text{or equivalently} \frac{dz}{dt}=\frac{x}{z}\frac{dx}{dt}.$$
Now freeze the situation $45$ seconds after the plane passed overhead. We can compute $x$ at that instant, and, using the Pythagorean Theorem, $z$, and now we know everything.
Note Many students would write instead that $z=\sqrt{1+x^2}$, and then differentiate. That's perfectly fine, a bit more work, a bit greater chance of error.
Best Answer
Let $A(t)$ be the area of the square at time $t$, let $\ell(t)$ be the length of the side of the square at time $t$.
We are given the following information: the rate at which the length of the side (that is, $\ell(t)$) changes is 6 units per second. This is a way of giving you information about the derivative of $\ell$. That is, it's telling you that the rate of change of $\ell$ with respect to time (why time? because it's "per second") is $6$ units per second. In symbols, it's telling you that" $$\frac{d\ell}{dt} = 6\text{ units/second}$$ We are asked to find how fast the area is changing when the area is 16 square units. "How fast" is a dead giveaway that they are asking you for the rate of change of the area; that is, they are asking you for the value of $\frac{dA}{dt}$, the derivative of the area, when the area is equal to 16. In symbols, they are asking you to find: $$\frac{dA}{dt}\Bigm|_{A=16}.$$
The first step, as in any related rates problem, is to find an equation that relates the quantities involved, in this case the length of the side $\ell$ and the area $A$. The relation between $A$ and $\ell$ is given by: $$A = \ell^2.$$ (Geometry: the area of a square is the square of the length of the side)
Once you have the relation between the original quantities, you take derivatives (implicitly) with respect to $t$ to obtain a relation between the quantities and their rates of change: $$\begin{align*} A &= \ell^2\\ \frac{d}{dt} (A) &= \frac{d}{dt}(\ell^2)\\ \frac{dA}{dt} &= 2\ell\frac{d\ell}{dt}. \end{align*}$$
Once you have a relation between the quantities and the rates of change, you plug in all the information you have and solve for the rate you want.
Since we want to know $\frac{dA}{dt}$, we need to know the values of $\ell$ and of $\frac{d\ell}{dt}$. The value of $\frac{d\ell}{dt}$ we already know: they told us from the beginning that it was always equal to $6$. So we just need to figure out the value of $\ell$ at the instant we are interested in.
The instant we are interested in is the instant when $A=16$. Since $A=\ell^2$, we need $16=\ell^2$; solving for $\ell$ and remembering that it is a length, so it must be nonnegative, we find that $16=\ell^2$ means that $\ell$ must be equal to $4$. Plugging in everything (the value of $\ell$, the value of $\frac{d\ell}{dt}$) into the formula we got that relates everything we get $$\begin{align*} \frac{dA}{dt} &=2\ell\frac{d\ell}{dt}\\ \frac{dA}{dt} &= 2(4)(6)\\ &= 48\text{ units}^2/\text{second}. \end{align*}$$ So at the instant that the area is 16 square units, the area is changing at a rate of 48 square units per second.