$$\lim_{x \to 0} \frac{(27+x)^{\frac{1}{3}} - 3}{9 - (27+x)^{\frac{2}{3}}}$$ equals:

We want to find the limit

$$\displaystyle \lim_{x \to 0} \frac{(27+x)^{\tfrac13}-3}{9-(27+x)^{\tfrac23}}.$$

First, set

$$A=(27+x)^{\tfrac13}.$$

Because taking the cube root and the square of the cube root will both be required, this single symbol will make every step clearer. Re-writing the original fraction in terms of $$A$$ we obtain

$$\frac{(27+x)^{\tfrac13}-3}{9-(27+x)^{\tfrac23}} =\frac{A-3}{\,9-A^{2}\,}.$$

Now use the algebraic identity for the difference of squares:

$$u^{2}-v^{2}=(u-v)(u+v).$$

With $$u=3$$ and $$v=A$$, this identity gives

$$9-A^{2}=(3)^{2}-A^{2}=(3-A)(3+A).$$

Notice that $$3-A=-(A-3)$$. Substituting this factorisation into the denominator we have

$$9-A^{2}=-(A-3)(3+A).$$

Placing this form in the fraction yields

$$\frac{A-3}{9-A^{2}} =\frac{A-3}{-(A-3)(3+A)} =\frac{A-3}{-1\cdot(A-3)(3+A)}.$$

Because $$A$$ tends to $$3$$ as $$x$$ tends to $$0$$, $$A-3$$ tends to $$0$$ but is not identically zero, so we can safely cancel the common factor $$A-3$$ appearing in the numerator and denominator:

$$\frac{A-3}{-1\cdot(A-3)(3+A)}=-\frac{1}{3+A}.$$

Now recall that $$A=(27+x)^{\tfrac13}$$. As $$x \to 0$$ we have $$A \to (27)^{\tfrac13}=3$$. Substituting $$A \to 3$$ into the simplified expression gives

$$\lim_{x \to 0}\left(-\frac{1}{3+A}\right) =-\frac{1}{3+3} =-\frac{1}{6}.$$

Hence, the correct answer is Option A.