Extremal values of $(x+3)^\frac{1}{3} - x^\frac{1}{3}$ $f(x) = (x+3)^\frac{1}{3} - x^\frac{1}{3}$

I am trying to find the extremal values of $f$. I start by differentiating, equating the derivative to 0, and solving that for x:

$f'(x)= \frac{1}{3} ((x+3)^\frac{-2}{3} - x^\frac{-2}{3}) = 0$

$(x+3)^2=x^2$

$x^2 + 6x + 9 = x^2$

$x = \frac{-3}{2}$

The extremal value of $f$ is at $f(\frac{-3}{2})$.

$f(\frac{-3}{2})= 2.289428485$ (decimal approximation)

However, when I try to verify this result with a CAS, WolframAlpha says something else, and SymboLab talks about the saddle of some interval, which I am not familiar with.

Is my working correct?
Your calculus is correct and $x=-\frac{3}{2}$ is a maximum as $f''(-\frac{3}{2})&lt;0$ as noted in the answer of @dewypeters.

But note that the first derivative $f'$ is not defined for $x=-3$ and $x=0$ that are points where the graph of the function has a vertical tangent.

The result of WolframAlpha may depend on this fact and, maybe, by the particular way in which you are write the question (that I cannot see, but you can see the answers to my question about the roots of the first derivative : A WolframAlpha error?). Anyway you can see here the graph of your function by WA.
I've got it.

$f''(\frac{-3}{2}) &lt; 0$, therefore the extremal is a maximum point.