Total Derivative at a point Let $f(x,y)=x^{3}+y^{3}$ .
How do I find the total derivative at (0,0) by using the definition?
I am confused of what to take as the Error function.

$$ (x+h)^3 +(y+k)^3 -x^3 -y^3 -3x^2 h -3y^2k = 3xh^2 +3yk^2 +h^3 +k^3 =\sqrt{h^2 +k^2} \left[(x+y)\sqrt{h^2 +k^2} +\frac{h^3 +k^3}{\sqrt{h^2 +k^2 }}\right] $$
setting 
$r(h,k) =(x+h)^3 +(y+k)^3 -x^3 -y^3 -3x^2 h -3y^2k$
$\phi (h,k )=(x+y)\sqrt{h^2 +k^2} +\frac{h^3 +k^3}{\sqrt{h^2 +k^2 }}$
we have $$\frac{|f((x,y) +(h,k)) -f(x,y) -[3x^2 ,3y^2 ]\circ [h,k] |}{||(h,k)||} =\frac{|r(h,k)|}{|| (h,k)||} =\phi (h,k) \to 0 $$ as $(h,k)\to (0,0).$
Thus $f'(x,y) =[3x^2 , 3y^2] .$
