Is $f(x,y)= \frac{x^3y^2-2y^4}{\sqrt{2x^2+y^4}}$ for $(x,y)\neq (0,0)$, $f(0,0)=0$, defining a differentiable function?
My attempt:
So I tried two approaches, either we can compute partial derivatives by definition of the derivation in the direction of canonical basis vectors, make a matrix out of them and try to show that that matrix is the gradient. Number two: calculate partial derivatives in $(x,y)\neq (0,0)$ and show that they're continuous in $(0,0)$ then the statement follows.
However, checking whether the partial derivatives are continuous at the point $(0,0)$ yields an even more difficult limit.
So I figured out that the potential candidate is the matrix consisting of two zeroes.
So, $$\lim_{(x,y) \rightarrow (0,0)} \frac{x^2y^2-2y^4}{\sqrt{(2x^4+y^4)(x^2+y^2)}}$$ but I'm not sure how to go about this. I think it is supposed to be differentiable, but I'm really not sure what to do about this. If it were not, I could just check for stuff like $\lim_{(x,x)->0}$ and some others, see that they're not the same thing, conclude that the function in this limit is not continuous at $(0,0)$ and that this limit doesn't exist.
Concerning these problems in general, is there anything better to do, when you suspect that the function is differentiable in $(0,0)$ other then trying to figure out such weird limits, because during a test I'm afraid I might not have enough time to figure out the way to calculate it.