Is the function $f(x,y) = \frac{xy^3}{x^4+y^2}$ a $C^1$-function? Consider the function 
$$
f(x,y) = \frac{xy^3}{x^4+y^2}.
$$
with $f(0,0) = 0$. 
Is this a $C^1$-function?
Firstly we compute: 
\begin{align*}
& D_1f(x,y) = \frac{y^3(x^4+y^2)-xy^3(4x^3)}{(x^4+y^2)^2} = \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2}, \\
& D_2f(x,y) = \frac{3xy^2(x^4+y^2)-xy^3(2y)}{(x^4+y^2)^2} = \frac{xy^2(3x^4+y^2)}{(x^4+y^2)^2}, 
\end{align*}
and $D_1f(0,0)= D_2f(0,0) = 0$. The function is differentiable at $(0,0)$ with $f'(\vec{0}) = [0,0]$, since we can verify that: 
\begin{align*}
\frac{|f(x,y)|}{\|\vec{x}\|} & = \frac{|x||y|^3}{x^4+y^2} \cdot \frac{1}{\sqrt{x^2+y^2}} = \frac{|x||y|}{\frac{x^4}{y^2} + 1} \cdot \frac{1}{\sqrt{x^2+y^2}} = \frac{|x||y|}{\left( \frac{x^2}{y} \right)^2 + 1} \cdot \frac{1}{\|\vec{x}\|} \\
& \leq |x||y| \cdot \frac{1}{\|\vec{x}\|} \leq \|\vec{x}\|^2 \cdot \frac{1}{\|\vec{x}\|} = \|\vec{x}\| \to 0 \textrm{ als } \vec{x} \to \vec{0}. 
\end{align*}
In order to determine if it is $C^1$, we need to determine whether the partial derivatives are continuous on the entire domain. I am trying to determine this for the point $(0,0)$, and so I need to determine
$$
\lim_{(x,y) \to (0,0)} \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2}, 
$$
and 
$$
\lim_{(x,y) \to (0,0)} \frac{xy^2(3x^4+y^2)}{(x^4+y^2)^2}. 
$$
I'm stuck in this step, I don't know how to compute these limits. 

You have
$$\left\vert \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2} \right\vert \le \vert y \vert^3\frac{(y^2+3x^4)}{(x^4+y^2)^2} \le 3 \vert y \vert^3 \frac{(y^2+x^4)}{(x^4+y^2)^2}= 3 \frac{\vert y \vert^3}{x^4+y^2} \le 3\vert y \vert$$ hence
$$\lim_{(x,y) \to (0,0)} \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2} = 0.$$ 
Using a similar argument you also get
$$\lim_{(x,y) \to (0,0)} \frac{xy^2(3x^4+y^2)}{(x^4+y^2)^2}=0.$$
This proves that the partial derivatives are continuous at the origin and that $f$ is $\mathcal C^1$.
