I can't figure out how the equation in yellow came from the step above.
$\text{Differentiate}\; (x+4)^{1/3}$
$\begin{align}\text{Solution:}\qquad \text{Let}\; &f(x)=(x+4)^{1/3}\qquad\qquad\qquad\qquad\qquad\qquad\qquad(1)\\ &f(x+\delta x)=(x+\delta x+4)^{1/3}\qquad\qquad\qquad\qquad\ \ \;\;\;(2)\end{align}$
$\begin{align}&\text{Subtracting (1) from (2), we get}\\ &f(x+\delta x)-f(x)=(x+\delta x+4)^{1/3}-(x+4)^{1/3}\\ &\qquad\qquad\qquad\;\;\;\;\ =(x+4+\delta x)^{1/3}-(x+4)^{1/3}\\ &\\ &\bbox[yellow,5px]{=\dfrac{\left[(x+4+\delta x)^{1/3}-(x+4)^{1/3}\right]\left[(x+4+\delta x)^{2/3}+(x+4+\delta x)^{1/3}(x+4)^{1/3}+(x+4)^{2/3}\right]}{(x+4+\delta x)^{2/3}+(x+4+\delta x)^{1/3}(x+4)^{1/3}+(x+4)^{2/3}}}\\ &\\ &=\dfrac{(x+4+\delta x)-(x+4)}{(x+4+\delta x)^{2/3}+(x+4+\delta x)^{1/3}(x+4)^{1/3}+(x+4)^{2/3}}\\ &\\ &=\dfrac{\delta x}{(x+4+\delta x)^{2/3}+(x+4+\delta x)^{1/3}(x+4)^{1/3}+(x+4)^{2/3}}\\ &\\ &\text{Dividing by}\; \delta x\; \text{and taking the limit as}\; \delta x\to0,\\ &\lim \limits_{\delta x \to 0} \dfrac{f(x+\delta x)-f(x)}{\delta x}=\lim \limits_{\delta x \to 0}\left[\dfrac{1}{(x+4+\delta x)^{2/3}+(x+4+\delta x)^{1/3}(x+4)^{1/3}+(x+4)^{2/3}}\right]\\ &\\ &f'(x)=\dfrac{1}{(x+4)^{2/3}+(x+4)^{1/3}(x+4)^{1/3}+(x+4)^{2/3}}=\dfrac{1}{3(x+4)^{2/3}}\end{align}$
Notice that everything in the brackets in the highlighted equation is exactly the same as the denominator of that massive fraction. It's the same as multiplying by $\frac11$, and it would seem that that particular expression was chosen to multiply on the top and bottom so as to end up with nothing but $\delta x$ in the numerator.