showing $f(x_1,x_2)=\sqrt[3]{x_1x_2}$ is differentiable Given $f\colon\mathbb R^{>0}\times\mathbb R^{>0}\rightarrow\mathbb R, (x_1,x_2)\mapsto \sqrt[3]{x_1\cdot x_2}$ I want to prove that $f$ is differentiable.
I know $f$ is partial differentiable and all partial derivatives are continuous, so $f$ is differentiable.
But I am wondering if I could show it by the definition of differentiability.
E.g. consider the point $1=(1,1)$. So there has to be a linear map $A$ such  that $\lim_{h\rightarrow0}\frac{f(1+h)-f(1)-Ah}{\|h\|}=0$
Let $h=(h_1,h_2)$. It's $A=(\frac13,\frac13)$ and so $\lim_{h\rightarrow0}\frac{\sqrt[3]{(1+h_1)(1+h_2)}-1-\frac13h_1-\frac13h_2}{\|h\|}=0$ ?
I know how to show it using polar coordinates but is there any way by doing it without any big 'help functions/relations' etc.?

Hint $\sqrt[3]{1+h_1}-1-\frac{h_1}{3}=O(h_1^2)$, then 
$$\sqrt[3]{1+h_1}\sqrt[3]{1+h_2}=(1+\frac{h_1}{3}+O(h_1^2))(1+\frac{h_2}{3}+O(h_1^2）)=1+\frac{h_1}{3}+\frac{h_2}{3}+O(h_1^2)+O(h_2^2).$$
To show $\sqrt[3]{1+h_1}-1-\frac{h_1}{3}=O(h_1^2)$, you may use the formula $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$.
