Proving derivabiliy of multivariable function. Is polar transformation helpful? Is $f(x,y)$ continuous and/or derivable at $(0,0)$.
$f(x,y)=\frac{x\sqrt{|x|}{y^2}}{x^2+y^4}$ when $(x,y)\neq(0,0)$.
$f(x,y)=0$ at $(x,y)=(0,0)$
Would converting this equation to polar form ($x=r \cosθ$ and $y^2=r\sinθ$) help? Why or why not? Is it even legible?
In my semester exam, it was told that polar form written above is not correct(we get the correct answer from it though), can someone clarify why? Can it be done by taking $x=r\sin θ,y=r\sin θ$ ?
 A: Converting to polar form is often helpful when dealing with problems of this kind, but here it isn't, for the following reason: The function behaves in an undesirable way only along the parabola $x=y^2$, which is not well captured in terms of polar coordinates.
Introduce the auxiliary function $\rho(x,y):=\sqrt{x^2+y^4}$. Then $x^2+y^4=\rho^2$ and therefore
$$|x|\leq\rho,\quad y^2\leq \rho\ .$$
It follows that for $(x,y)\ne(0,0)$ one has
$$\bigl|f(x,y)\bigr|={|x|^{3/2} y^2\over \rho^2}\leq{\rho^{5/2}\over\rho^2}=\sqrt{\rho(x,y)}\ .$$
It follows that $\lim_{(x,y)\to(0,0)}f(x,y)=0$, hence $f$ is continuous at $(0,0)$.
But $f$ is not differentiable at $(0,0)$. 
Proof. If $f$ were differentiable at $(0,0)$ then the composition of $$t\mapsto\bigl(x(t),y(t)\bigr):=(t^2,t)$$
with $f$would have to be differentiable at $t=0$. But
$$t\mapsto f\bigl(x(t),y(t)\bigr)={t^2\>|t|\>t^2\over 2t^4}={|t|\over2}$$
is not differentiable at $t=0$.$\quad\square$
A: Edit: I've actually rethought the problem, and found solution "with" polar coordinates. For a simple solution check the last line.

Christian already gave a beautiful solution to the problem, but I'll discuss why polar coordinates are not that helpful in this case.
First let me explain how we use polar coordinates in general to find limit at $(0,0)$. When you convert to polar coordinates you get some $f(r,\varphi)$ and you want to find some function $g$ such that $|f(r,\varphi)-L| \leq |g(r)|$ no matter what $\varphi$ is and such that $g(r)\to 0$ when $r\to 0$. Usual tricks involve something like $|\sin\varphi|,|\cos\varphi|\leq 1$ to get rid of $\varphi$ from $f(r,\varphi)$. 
In this case, we want to prove that the limit is $0$, so plugging the polar coordinates in $f$ gives us
$$\left| \frac{r\cos\varphi\sin^2\varphi\sqrt{r|\cos\varphi|}}{\cos^2\varphi + r^2\sin^4\varphi} \right|$$ 
and it is not at all obvious how to make this work.
One idea that occurred to me is if we could have $\cos^2\varphi + r^2\sin^2\varphi\geq r$ when $r$ is near $0$, then the above expression could be bounded as
$$\left| \frac{r^{3/2}\cos\varphi\sin^2\varphi\sqrt{|\cos\varphi|}}{\cos^2\varphi + r^2\sin^4\varphi} \right|\leq \left| \frac{r^{3/2}}{r} \right|\leq \sqrt r$$
but this doesn't work for, say, $\varphi = \pi/2$.
However, there is a way to handle this using the following inequality:
$$\left|\frac{xy}{x^2+y^2}\right|\leq \frac 12\tag{*}$$ (which is implied by $(|x|-|y|)^2\geq 0$) in the following form:
$$\left| \frac{r\cos\varphi\sin^2\varphi\sqrt{r|\cos\varphi|}}{\cos^2\varphi + r^2\sin^4\varphi} \right| = \sqrt {r|\cos\varphi|}\left| \frac{\cos\varphi(r\sin^2\varphi)}{\cos^2\varphi + (r\sin^2\varphi)^2} \right|\leq \frac{\sqrt r}2$$ and we are done.

Everything is nice, but... Did we essentially use polar coordinates? I mean, did we benefit from writing all of this? Not really, and here is why:
$$\left| \frac{x\sqrt{|x|}y^2}{x^2+y^4} \right| = \sqrt{|x|}\left|\frac{xy^2}{x^2+(y^2)^2}\right|\stackrel{(*)}\leq\frac{\sqrt{|x|}}{2}.$$
