## Calculus 8th Edition

$1$
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 $\lim\limits_{h \to 0}\dfrac{h}{h}=\dfrac{\partial f }{\partial x}(0,0)=1$