Total derivative proof Let $f(x,y) = \frac{xy^2}{|x| +y^2}$ and $f(0,0)=0$ show that $f$ is total differentiable at $(0,0)$ using the definition of total differentiation . Any hints about using the definition or how the definition works here ? 
 A: Using vector notation:
The function $f(\mathbf x)$ is differentiable at $ \mathbf x $ if you can approximate it in a neighborhood of $\mathbf x$ as follows:
$$ f(\mathbf x+\mathbf h) = f(\mathbf x) + f_{\mathbf x}^{'}(\mathbf h)+r(\mathbf h) $$
For this, the differential $f_{\mathbf x}^{'} $ has to exist at every point in the neighborhood and 
$$ \lim_{\mathbf h \to \mathbf 0}\frac{r(\mathbf h)}{|\mathbf h|}= 0 $$
Explicitly, in this case, let $ \mathbf h = (h_1,h_2)$ 
$$ f_{\mathbf x}^{'}(\mathbf h) = \frac{\partial f}{\partial x}h_1+\frac{\partial f}{\partial y}h_2 $$
with the partials being evaluated at $(0,0)$.
Using first principles, we can evaluate at $(0,0)$:
$$ f_{\mathbf x}^{'}(\mathbf h) = 0 $$
Plugging this back into the original equation,
$$ r(\mathbf h) = \frac{h_1 h_2^2}{|h_1|+h_2^2} $$
$$\frac{r(\mathbf h)}{|\mathbf h|} =\frac{h_1 h_2^2}{(|h_1|+h_2^2)|\mathbf h|} $$
Let 
$$ g = g(h_1,h_2) = \frac{h_2}{h_1} $$ 
$$\frac{r(\mathbf h)}{|\mathbf h|} =\frac{h_2}{(\pm 1+h_2 \cdot g) \sqrt{1+g^2}} $$
Finally, note that this expression is zero in the limit $ |\mathbf h| \to 0$, whether $g$ is finite or unbounded. 
