Here is an answer that starts from the tangent line approximation, in order to visualize the differentials approach better. Let $f(x)=x^{1/3}$. Then $f'(x)=\frac{1}{3}x^{-2/3}$.
The point $x=27$ is very near to $27.05$, and our function is easy to evaluate at $27$, since $27^{1/3}=3$. Let us find the equation of the tangent line to $y=x^{1/3}$ at $x=27$. The slope is $(1/3)27^{-2/3}=1/27$. So the equation of the tangent line is $y-3=(1/27)(x-27),\tag{$\ast$}$ or equivalently $y=\frac{x}{27}+2.$ Recall that the tangent line at $x=27$ kisses the curve at $x=27$. So near $x=27$, the tangent line is very close to the curve. Now let $x=27.05$. The $y$-coordinate of the point on the curve with $x$-coordinate $27.05$ is given approximately by $\frac{27.05}{27}+2=3+\frac{0.5}{27}.$ That is the desired approximation.
For differentials, it is best to go back to $(\ast)$.
Let $\Delta x$ be the change in $x$. In our case, $\Delta x=0.05$. Let $\Delta y$ be the corresponding change in $y=f(x)$ as we move from $x=27$ to $x=27+\Delta x$. Then from $(\ast)$ we have $\Delta y\approx \frac{1}{27}\Delta x$.
More casually, $y+dy\approx y +\frac{dy}{dx}dx$. We take $dx=0.05$. We found that at $x=27$, $\frac{dy}{dx}=\frac{1}{27}$. So $y+dy\approx 3+\frac{1}{27}(0.05)$.
Exactly how you should write out the calculation depends on the notation that your instructor uses.