Multivariable limit - perhaps a trickier problem I am stuck on. I am trying to solve the following limit:
$\lim_{(x,y) \to (0,0)} \frac{x^4y^4}{(x^2 + y^4)^3}$
(This is a more challenging problem from Folland Calculus, it seems).
I am pretty sure this limit does not exist (however, this is just a guess, and I am not 100% sure.)
I am trying to approach via the paths y = mx and x = my but... they don't seem to work. I have tried simpler cases such as x = 0 and y = 0. 
Any help appreciate.

Use polar coordinates :
$$\frac{x^4y^4}{(x^2+y^4)^3} = \frac{r^8\cos^4\theta\sin^4\theta}{r^6(\cos^2\theta+r^2\sin^4\theta)^3} = r^2\frac{\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3}$$
Now the problem remaining is : is the fraction bounded or not ? You have, for $\theta\ne\frac\pi2\mod\pi$ :
$$\frac{\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3} = \frac{\cos^4\theta\sin^4\theta}{\cos^6\theta\left(1+r^2\frac{\sin^4\theta}{\cos^2\theta}\right)^4}$$
So if you go to $0$ following the curve of equation $r=\cos\theta$, for example (this is a circle tangent to the $y$ axis at the origin), you have :
$$\frac{x^4y^4}{(x^2+y^4)^3} = \cos^2\theta\frac{\cos^4\theta\sin^4\theta}{\cos^6\theta(1+\sin^4\theta)}\xrightarrow[\theta\to\frac\pi2]{}\frac12\ne0$$
So your function is not continuous at the origin, as you suspected.
