Small Changes of Sphere 
Attacking the question using this formula:
$$f(x_0+h)\approx f(x_0)+hf'(x_0)\\
f(r)\approx \frac43\pi r^3\\
f'(r)\approx 4\pi r^2\\
\text{let }x_0=0,h=\Delta r\\
$$
This doesn't seem to work as now both fuctions will equate to zero, so I tried :
$$\dfrac{dV}{dr}\approx \frac{\Delta V}{\Delta r}\\
4\pi r^2\approx \frac{\Delta V}{\Delta r}\\
\Delta V\approx 4\pi r^2\Delta r
$$
Which is not the same as the question.
 A: As David notes in the comments, $\Delta V$ should have units of length cubed, so the given answer is probably a typo.  For an alternate method, you can always just attack it directly.  To wit:
Let's say $\Delta r = \delta$.  Letting $V'$ denote the new volume,  $V' = \displaystyle \frac{4}{3} \pi (r+\delta)^3$.  Expanding this out:
$$V' = \frac{4}{3} \pi \Big( r^3 + 3\delta r^2 + 3\delta^2r + \delta^3 \Big)$$
Since $\delta$ is very small, we can use the approximation $\delta^n = 0$ for all $n > 1$.  So this becomes:
$$V' \approx \frac{4}{3} \pi \Big( r^3\ + 3\delta r^2 \Big) = V + 4 \pi \delta r^2$$
The difference in volume, $\Delta V = V' \!-\! V$, is therefore roughly $4 \pi \delta r^2 = 4 \pi r^2 ( \Delta r )$.  
A: You are right with your $\Delta V \approx 4\pi r^2\Delta r$
You can be certain that the equation $\Delta V \approx 4\pi r^2\Delta r^2$ of the question has a typo as it is not homogeneous. More precisely, on the left side the dimension is $L^3$ but on the right side it is $L^4$, hence it is wrong for sure (where $L$ is a length). It is always a good reflex to check homogeneity, you can easily control your (and others!) answers and know if it is false for sure :). 
A: There are 2 problems with your initial approach. 
 1) $x_0 \neq 0$. Actually $x_0 = r$ since the initial radius of the sphere is not zero.
 2) You may understand this but, your equation doesn't just describe $\Delta V$. Your first order Taylor approximation can also be written as $V=V_0 + \Delta V$. Thus: $$\Delta V = V-V_0=f(x_0+h)-f(x_0)=hf'(x_0)=\Delta r \cdot 4 \pi r^2$$
