Is the function $f(z) = \mathfrak{Re}(z) \mathfrak{Im}(z)$ differentiable at $z=0$? I am trying to approach the problem by checking whether or not the limit 
$$\lim_{z\to 0} \frac{f(z + 0) - f(0)}{z}$$
exists. If $z = x+iy$, then the expression in the limit becomes
$$ \frac{x^2y}{x^2 + y^2} - \frac{xy^2}{x^2+y^2}\,i.$$
Now what I usually do is to try and approach the limit from two different paths to get different results (and therefore the limit does not exist).  When this cannot be done, and I start to suspect that the limit exists, then I try to manipulate the expression so that I can evaluate it directly. 
However this technique isn't being very fruitful for this particular example. How can I determine whether $f(z) = \mathfrak{Re}(z) \mathfrak{Im}(z)$ differentiable at $z=0$?

Note: By differentiable here I mean pointwise differentiability, not differentiability in the complex sense (i.e. holomorphicity).
 A: Presumably, you've figured out, by testing paths, that the limit must be zero if it exists. This basically boils down to testing that
$\frac{x^2y}{x^2+y^2}$ and $\frac{xy^2}{x^2+y^2}$ actually go to zero when both $x$ and $y$ do. There's sort of no way around making actual estimates in a case like this - however, we can use the intuition that the denominator has all terms degree $2$ and the numerator has all terms degree $3$ to work this out.
In particular, note that $|x^2y| \leq (x^2+y^2)^{3/2}$ since $x$ and $y$ are both bounded by $(x^2+y^2)^{1/2}$. In particular, dividing both sides by $x^2+y^2$ gives $\left|\frac{x^2y}{x^2+y^2}\right|\leq (x^2+y^2)^{1/2}$ and clearly $(x^2+y^2)^{1/2}$ goes to zero as $x$ and $y$ do - thus $\lim_{(x,y)\rightarrow (0,0)}\frac{x^2y}{x^2+y^2}=0$. Similar reasoning handles the imaginary part.
One can also apply this reasoning directly on a complex number $z$: Clearly $|\Im(z)|$ and $|\Re(z)|$ are bounded by $|z|$. Thus, their product is bounded by $|z^2|$, so $\left|\frac{\Re(z)\Im(z)}z\right|\leq |z|$, which obviously goes to zero.
So, this function is indeed differentiable at zero.
(Though, to avoid confusion, most things in complex analysis assume differentiability on an open set. Being differentiable at a point doesn't have any particularly nice implications. This function is not differentiable anywhere else)
A: With $z=x+iy,$
$$\left |\frac{f(z) - f(0)}{z-0}\right | = \left |\frac{f(z)}{z}\right |= \frac{|x||y|}{|z|} \le \frac{|z|^2}{|z|} = |z|\to 0.$$
It follows that $f'(0)=0.$
