Find first derivative of a function $f(x) = x\sqrt[3]{x}$ using definition

I'm having trouble finding first derivative of a function:

$f(x) = x\sqrt[3]{x}$

using definition:

$\displaystyle \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

I should get:

$\displaystyle \lim_{h \to 0} \frac{(x+h)(\sqrt[3]{x+h})-x\sqrt[3]{x}}{h} $

But I don't know what my next step should be.

Could somebody give me a hint please?

Hint: $\dfrac{(x+h)\sqrt[3]{x+h} - x\sqrt[3]{x}}{h}= x\cdot\dfrac{\sqrt[3]{x+h} - \sqrt[3]{x}}{h}+\sqrt[3]{x+h}$. From this you can simplify the factor next to $x$ of the first term using the identity: $a - b = \dfrac{a^3 - b^3}{a^2+ab+b^2}$

If you multiply both the numerator and the denominator of that fraction by$$(x+h)^2\sqrt[3]{x+h}^2+(x+h)\sqrt{x+h}\,x\sqrt[3]x+x^2\sqrt[3]x^2,$$then the limit becomes\begin{multline}\lim_{h\to0}\frac{(x+h)^3(x+h)-x^4}{h\bigl((x+h)^2\sqrt[3]{x+h}^2+(x+h)\sqrt{x+h}\,x\sqrt[3]x+x^2\sqrt[3]x^2\bigr)}=\\=\lim_{h\to0}\frac{6hx^2+4h^2x+h^3+4x^3}{(x+h)^2\sqrt[3]{x+h}^2+(x+h)\sqrt{x+h}\,x\sqrt[3]x+x^2\sqrt[3]x^2}.\end{multline}Can you take it from here?