Solve $\lim\limits_{n\to\infty}\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}$

$\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=\frac{n+\sqrt{n}-n}{(n+\sqrt{n})^{2/3}+(n+\sqrt{n})^{1/3}n^{1/3}+n^{2/3}}\leq \frac{\sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=\frac{\sqrt{n}}{3n^{2/3}}\rightarrow 0$

Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$

We have that $$0< \sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=\frac{\sqrt[3]{n}}{\sqrt{n}}\cdot\frac{ \sqrt[3]{1+\frac{1}{\sqrt{n}}}-1}{\frac{1}{\sqrt{n}}}\leq \frac{1}{\sqrt[6]{n}}\cdot \frac{1}{3}.$$ where we used the Bernoulli's inequality $$(1+x)^r\leq 1+rx$$ with $r=1/3\in(0,1)$ and $x=1/\sqrt{n}$.

Can you take it from here?

P.S. By using this approach we are also able to find
$$\lim\limits_{n\to\infty}\left((n+\sqrt{n})^r-n^r\right)$$ for any $r<1/2$. Note that if we replace $\sqrt[3]{\ }$ with say $\sqrt[5]{\ }$ then the "algebraic way" could be very annoying!

By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have

$$\sqrt[3]{n+\sqrt{n}}=\sqrt[3]{n}\, \left(1+\frac1{\sqrt n}\right)^\frac13=\sqrt[3]{n}+\frac{\sqrt[3]{n}}{3\sqrt n}+o\left(\frac{\sqrt[3]{n}}{\sqrt n}\right)=\sqrt[3]{n}+\frac{1}{3n^\frac16}+o\left(\frac{1}{n^\frac16}\right)$$

therefore

$$\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=\frac{1}{3n^\frac16}+o\left(\frac{1}{n^\frac16}\right)$$