Related rates - approximation or not? 
The radius of a uniform spherical balloon is increasing at 3% per
  second. Find the % rate at which its volume is increasing.

My solution:
Percentage is change of some value divided by that value so the percentage rate at which the volume is increasing will be given by:
$$\frac{\frac{dv}{V}}{dt}$$
We already know that $$\frac{\frac{dr}{r}}{dt}=3\frac{\%}{s}$$
Then:
$$\frac{\frac{dV}{V}}{dt} = \frac{\frac{dV}{V}}{\frac{dr}{r}}* \frac{\frac{dr}{r}}{dt}$$
$$\frac{\frac{dV}{V}}{dt} = \frac{dV}{V}* \frac{r}{dr} * 3$$
$$\frac{\frac{dV}{V}}{dt} = \frac{dV}{dr}* \frac{r}{V} * 3$$
$$\frac{\frac{dV}{V}}{dt} = 4\pi r^{2}* \frac{r}{V} * 3$$
$$\frac{\frac{dV}{V}}{dt} = \frac{4\pi r^{3}}{\frac{4}{3}\pi r^{3}} * 3$$
$$\frac{\frac{dV}{V}}{dt} = \frac{4\pi r^{3}}{\frac{4}{3}\pi r^{3}} * 3$$
$$\frac{\frac{dV}{V}}{dt} = 3 * 3$$
$$\frac{\frac{dV}{V}}{dt} = 9\frac{\%}{s}$$
Questions:
1) Is such reasoning correct? I'm particularly concerned about this transformation:
$$\frac{\frac{dV}{V}}{\frac{dr}{r}}=\frac{dV}{dr}* \frac{r}{V}$$
I remember reading that I can't treat derivatives as fractions but it happens to work here and this confuses me.
2) Is the final result exactly 9 % per second or is this only an approximation? For example the answer to:

If $V=2x^{3}$ what is the approximate percentage change in $V$ when
  $x$ changes by 2%?

is 6% but this is only an approximation (see Is dv only approximate of dv/dx*dx?) Is this the case in the original exercise or is 9 % per second an exact result?
 A: The actual percentage increase of the volume is $100\cdot(1.03^3-1)\% = 9.2727\%$, so your result of $9\%$ is only an approximation.
You could rephrase your problem just like the one at the bottom of your post:

If $V = \frac{4}{3} \pi r^3$ and $r$ increases by $3\%$, what is the $\%$ increase in $V$?

Well if
$$
V_\text{old} = \frac{4}{3} \pi r_\text{old}^3
$$
and $r_\text{new} = 1.03r_\text{old}$, then
$$
\begin{align}
V_\text{new} &= \frac{4}{3}\pi r_\text{new}^3 \\
    &= \frac{4}{3} \pi r_\text{old}^3 \cdot 1.03^3 \\
    &= 1.03^3 V_\text{old}.
\end{align}
$$
So, $V$ increased by a factor of $1.03^3$, and hence its percentage increase is $100\cdot(1.03^3-1)\%$.

The reason your method gives the wrong answer is that
$$
\frac{\frac{dr}{r}}{dt}=3\frac{\%}{s}
$$
should actually be
$$
\frac{\frac{dr}{r}}{dt}=\log(1.03).
$$
Note that $\log(1.03) \approx 0.03 = 3\%$ because $\log(1+x) \approx x$ for small $x$.  This is why your answer is approximately correct.
If the percentage increase in $r$ was much larger than $3\%$, say $50\%$, your approximation would be much worse: your method would say that $V$ would increase by $150\%$, while the actual percentage increase would be $237.5\%$.
Why the logarithm? Well, you know that $r$ is increasing proportionally to its size, i.e.
$$
\frac{dr}{dt} = cr
$$
for some constant $c$.  We can solve this equation and get
$$
r(t) = r_0 e^{ct}.
$$
To find out what $c$ should be we use the assumption that $r$ increases by $3\%$.  After $1$ second, $r$ should be equal to $3\%$ more than $r_0$, and this is $1.03r_0$. Plugging this in,
$$
\begin{align}
r(1) &= r_0 e^c \\
1.03 r_0 &= r_0 e^c,
\end{align}
$$
and cancelling $r_0$ from both sides gives
$$
1.03 = e^c,
$$
so $c = \log(1.03)$.  Thus
$$
\frac{dr/dt}{r} = \log(1.03).
$$
This should be your starting point.

Now, the reason everything you write after this works (i.e. why treating differentials like fractions works here) is that there are hidden "$/dt$"s floating around underneath everything that you have simply neglected to write. So, while your method is essentially correct, the way you have written it is not. The equation
$$
\frac{\frac{dV}{V}}{\frac{dr}{r}}=\frac{dV}{dr}* \frac{r}{V},
$$
for instance, makes no sense. Here's how your calculation should be written:
$$
\frac{dr}{dt} = r\log(1.03), \tag{1}
$$
so
$$
\begin{align}
\frac{dV}{dt} &= \frac{dV}{dr} \frac{dr}{dt} \qquad \text{(chain rule)} \\
&= 4\pi r^2 \frac{dr}{dt} \\
&= 4\pi r^3 \log(1.03) \\
&= V 3\log(1.03). \tag{2}
\end{align}
$$
Now there are two ways we can get back to the percentage increase. The first is to notice that we got the coefficient $\log(1.03)$ in equation $(1)$ by taking the log of $103\%$, so if we want a percentage from $3\log(1.03)$ then we should do the opposite and exponentiate it, giving
$$
e^{3\log(1.03)}\% = 1.03^3\% = 109.2727\%, \tag{3}
$$
which represents an increase of $9.2727\%$.
The second way we could obtain this answer is by solving the differential equation $(2)$ for $V$, obtaining
$$
V(t) = V_0 e^{3\log(1.03)t} = V_0 (1.03)^{3t}.
$$
The initial size of $V$ is $V(0) = V_0$, and after $1$ second its size is
$$
V(1) = V_0 1.03^3,
$$
so it has increased by a factor of $1.03^3 = 1.092727$, which is an increase of
$$
100\cdot(1.03^3 - 1)\% = 9.2727\%.
$$
