What are the critical values, if any, of f(x)= (2-x)/(x+2)^3?

A critical point on a function $f \left(x\right)$ occurs at ${x}_{o}$ if and only if either $f ' \left({x}_{o}\right)$ is zero or the derivative doesn't exist.

Hence we need to compute the first derivative

$\mathrm{df} \frac{x}{\mathrm{dx}} = d \frac{\frac{x - 2}{x + 2} ^ 3}{\mathrm{dx}} = \frac{x - 4}{x + 2} ^ 3$

which nullifies at ${x}_{o} = 4$

Using the second derivative test for local extrema we need to compute the second derivative for $f \left(x\right)$ hence we have that

${d}^{2} f \frac{x}{{d}^{2} x} = - 6 \cdot \frac{x - 6}{x + 2} ^ 5$

Because $f ' ' \left(4\right) > 0$ hence the function has a minimum at $x = 4$
which is $f \left(4\right) = - \frac{1}{108}$

It does not have a maximum as

${\lim}_{x \to - 2} f \left(x\right) = \infty$