In the limit question we have to calculate
$$\lim_{x \rightarrow 0^+ } \dfrac{\sqrt[3]{x+3x^2}-\sqrt[3]{x}}{\sqrt[3]{x^4}}$$
Can I use the method of equivalent infinitesimal to replace the functions?
Like we replace $\sin x $ with $x$ when $x \rightarrow 0 $.
In the question removing the cube roots doesn't remove the indeterminacy. So, can I remove the cube roots?
I want to know the deep maths behind this method. I only use this method to solve limits fast but I don't know whether this is applicable at all places?
You can simplify the expression in your problem a little further:
\begin{align*} \frac{(x+3x^2)^{1/3}-x^{1/3}}{x^{4/3}}=\frac{x^{1/3}\Big((1+3x)^{1/3}-1\Big)}{x^{4/3}}=\frac{(1+3x)^{1/3}-1}{x} \end{align*}
Then you can use the infinitesimal equivalency that Robert Z mentioned in his comment above: $(1+3x)^{1/3}\stackrel{\cdot}{\sim}1+x$