Show that $f(x,y) = (x^2y-\frac13y^3, \frac13x^3-xy^2)$ is totally differentiable and calculate its derivative. I'm working on the problem 

Show that $f(x,y) = (x^2y-\frac13y^3, \frac13x^3-xy^2)$ is totally differentiable and calculate its derivative.

And arived at the answer $Df(x,y)=(x^2+2xy-y^2, x^2-2xy-y^2)$. But I have doubts about the method I used to get there so I'd love someone to verify my work. (In specific, I'm not entirely certain I'm allowed to just casually ignore all terms with an $h$ in them.)
Solution: Consider $$\lim_{h\rightarrow 0} \frac{|f(x+h, y+h) - f(x,y) - A(h,h)|}{|(h,h)|}=0.$$ For conveinience, we'll focus our attention to the enumerator of this fraction and leave out the limit. (Leaving out a few algebraic steps for conciseness) We can then write this as $$\left|(x^2y+2xyh+yh^2+hx^2+h^3-\frac13y^3-y^2h-yh^2-\frac13h^3, \frac13 x^3+x^2h+xh^2+h^3-xy^2-2xyh-xh^2-hy^2-2yh^2-h^3) - (x^2y-\frac13y^3, \frac13x^3-xy^2)-A(h,h)\right|.$$ Note how all terms which are not multiplied by $h$ at least once cancel. This means that we can write this as $$\left|(h,h)\right|\left|(2xy+yh+x^2+h^2-y^2-yh-\frac13h^2, x^2+h^2-2xy-y^2-2yh-h^2)-
 A\right|$$
But remembering our limit, we know that we're going to have to divide by $|(h,h)|$ and that $h\rightarrow 0$ so that we have that $$|(2xy+x^2-y^2, x^2-2xy-y^2) - A| = 0,$$ so $A = (x^2+2xy-y^2, x^2-2xy-y^2)$.
Another question: The place I got this from says that the problem should take anywhere from 10 to 15 minutes to solve, but with this method it takes longer. What alternative methods should I consider in solving these kinds of problems, or what should my first line of attack be? 
For context, another they had listed within the same time frame was:

Show that $f(x,y) = |x+\pi|^3e^{3y}$ is totally differentiable and calculate its gradient.

 A: One of the most useful theorems in multivariate analysis is the following: If a function  $f:\>{\mathbb R}^n\to{\mathbb R}^m$ has continuous first partial derivatives then it is differentiable. Furthermore at each point ${\bf x}$ the matrix of $df({\bf x})$ is the matrix of these $m\times n$ partial derivatives. In your example the components of $f$ are polynomials in the coordinate functions $x$, $y$. Therefore the partial derivatives exist and are continuous in all of ${\mathbb R}^2$, and we obtain
$$\bigl[df(x,y)\bigr]=\left[\matrix{2xy&x^2-y^2\cr x^2-y^2&-2xy\cr}\right]\ .$$
To prove differentiability of $f$ from scratch we would have to compute
$$f(x+X,y+Y)-f(x,y)=\ldots=:\Phi(x,y,X,Y)$$
and sort terms according to increasing total degree in $X$ and $Y$. There will be no term of degree $0$, then terms of degree $1$ (which are the interesting ones), and a large number of higher degree terms. The latter are all of order $o\bigl(\sqrt{X^2+Y^2}\bigr)$ when $(X,Y)\to(0,0)$, as required in the definition of differentiability. (Note that $x$ and $y$ are constant in this argument.) This cannot be accomplished in $15$ minutes.
A: The function in question can be written as $$f(x,y)=x^2y-{1\over3}y^3+i(xy^2-{1\over3}x^3)$$
by rearranging terms and substituting $z=x+iy$ and $iz=ix-y$ we obtain:
$$f(x,y)=xy(x+iy)-{1\over 3}(y^3+ix^3)=xyz-{1\over 3}(y-ix)(y^2-x^2+xyi)=xyz+{1\over 3}iz(-z^2+3xyi)=-{1\over 3}iz^3$$
Then the function has been explicitly described as of z and its derivation is:
$$f^{'}(z)=-iz^2$$
