$f(x)=\sqrt[3]{x^2}$ derivative by defintion 
Find $f(x)=\sqrt[3]{x^2}$ derivative by definition


my first try was
$\lim_\limits{h \to 0}\frac{\sqrt[3]{(x+h)^2}-\sqrt[3]{x^2}}{h}$
I tried multiplying by $\frac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x^2}}$ and got$\lim_\limits{h \to 0}\frac{(x+h)(\sqrt[3]{x+h})-x\sqrt[3]x}{h\cdot(\sqrt[3]{(x+h)^2}+\sqrt[3]{x^2})}$
but I don't know how to continue from here
I also tried multiplying by $\frac{(x+h)^{1/3}+x^{1/3}}{(x+h)^{1/3}+x^{1/3}}$  but this seemed more confusing
Also tried writing $\sqrt[3]{(x+h)^2}+\sqrt[3]{x^2}$ as $(\sqrt[3]{(x+h)}-\sqrt[3]x)\cdot(\sqrt[3]{x+h}+\sqrt[3]x$ but this also did not help..
thanks for any help and tips

First of all, I would suggest you to apply the following identity:
\begin{align*}
a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})
\end{align*}
According to it, one concludes the desired result:
\begin{align*}
f'(x_{0}) & = \lim_{x\to x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}}\\
& = \lim_{x\to x_{0}}\frac{x^{2/3} - x^{2/3}_{0}}{x - x_{0}}\\
& = \lim_{x\to x_{0}}\frac{x^{2} - x^{2}_{0}}{(x - x_{0})(x^{4/3} + x^{2/3}x^{2/3}_{0} + x^{4/3}_{0})}\\
& = \lim_{x\to x_{0}}\frac{x + x_{0}}{x^{4/3} + x^{2/3}x^{2/3}_{0} + x^{4/3}_{0}}\\
& = \frac{2x_{0}}{3x^{4/3}_{0}} = \frac{2}{3x^{1/3}_{0}}
\end{align*}
Hopefully this helps!
