Answer
$1$
Work Step by Step
The derivative of a function at the some point can be defined as:
$\lim\limits_{h \to 0}\dfrac{f(x+h,y)-f(x,y)}{h}=\dfrac{\partial f }{\partial x}$
At $(x,y)=(0,0)$
$\lim\limits_{h \to 0}\dfrac{f(0+h,0)-f(0,0)}{h}=\dfrac{\partial f }{\partial x}(0,0)$
$\lim\limits_{h \to 0}\dfrac{\sqrt [3] {h^3+0^2}-\sqrt [3] {0^3+0^2}}{h}=\dfrac{\partial f }{\partial x}(0,0)$
After solving the above equations, we have
$\lim\limits_{h \to 0}\dfrac{h}{h}=\dfrac{\partial f }{\partial x}(0,0)=1$
