The function is:
$$ f(x,y)=\begin{cases} xy^3 \sin \left(\frac{1}{{x^4+y^4}}\right), & \text{if } (x,y) \neq (0,0) \\ 0, & \text{if } (x,y) = (0,0) \end{cases}.$$
I have to check if it is of class $C^1$.
I've found that
$d_xf(0,0) = 0$
and
$d_{y} f(0,0) = 0$
I know how to prove by definition it's differentiable at the origin and obviously it's differentiable elsewhere. But I don't know how to prove that partial derivatives are/are not continuous. For example i get
$d_xf(x,y) = y^3\sin\left(\frac{1}{x^4+y^4}\right) - \frac{(4x^4y^3)\cos\left(\frac{1}{x^4+y^4}\right)}{(x^4+y^4)^2}$
Any kind of help would be appreciated.
We have that $f_x(0,0)=0$ but $\lim_{(x,y)\rightarrow (0,0)} f(x,y) \neq 0$ indeed for $x=y=t \to 0$
$$f_x(x,y)= y^3\sin\left(\frac{1}{x^4+y^4}\right) - \frac{(4x^4y^3)\cos\left(\frac{1}{x^4+y^4}\right)}{(x^4+y^4)^2}=t^3\sin\left(\frac{1}{2t^4}\right) - \frac{\cos\left(\frac{1}{2t^4}\right)}{t}$$
with $t^3\sin\left(\frac{1}{2t^4}\right)\to 0$ but $\frac{\cos\left(\frac{1}{2t^4}\right)}{t}$ clearly doesn't admit a limit.
To show that let consider for example $\frac{1}{2t_n^4}=2\pi n \implies t_n=\frac1{\sqrt [4]{4\pi n}} $ such that
$$\frac{\cos\left(\frac{1}{2t_n^4}\right)}{t_n}=\sqrt [4]{4\pi n} \to \infty$$
Therefore the function is not of class $C^1$.