Show that the partial derivatives of $f$ are defined when $(x,y)=(0,0)$
for $f(x,y)=\frac {2x^2y^3}{x^4+y^6}$ when $(x,y)\ne (0,0)$, and $f(x,y)=0$ when $(x,y)=(0,0)$
I don't know how to use the limit definition of the partial derivatives to show this...
I have $\dfrac{\partial f(x,y)} {\partial x}=lim_{h\to 0} \frac{f(x+h,y)-f(x,y)}{h}$. Where do I go from here?
$\dfrac{\partial f(0,0)} {\partial x} = lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h}$
Is that correct?