Calculate the partial derivatives at $(0,0)$

\begin{equation} f(x,y)=\frac{x^2\sin{y^2}}{x^2+y^4} \text{ if }(x,y) \neq (0,0) \text{ and } f(0,0)=0 \end{equation}

Calculate the partial derivatives in $(0,0)$. Then show that $f$ isn't continuously differentiable in $(0,0)$.

I used the definition of derivatives to show that:

\begin{equation} \frac{\partial f}{\partial x}(0,0)= \frac{\partial f}{\partial y}(0,0)=0 \end{equation}

But how do I show this without using the definition? Or is it not possible in this case to calculate the partial derivatives using simple differentation rules? Also, since the function always equals zero if either $x$ or $y$ equals zero, the partial derivatives will also equal zero and zero is a continuous function for all values. So that would make it a continuously differentiable function, yet it isn't. Where am I going wrong?

Since the function is defined piecewise, you can't use the simple differentiation rules to check whether it is differentiable at $(x,y) = (0,0)$. At $(x,y) \neq 0$, we can use the regular differentiation rules and obtain for example

$$g(x,y) = \frac{\partial f}{\partial x} = \frac{2 x \sin y^2 (x^2 + y^4) - 2x^3 \sin y^2}{(x^2 + y^4)^2} = \sin y^2 \frac{2xy^2}{(x^2 + y^4)^2}.$$

To check whether $f$ is continuously differentiable at $(x,y) = (0,0)$ we need to check whether $\lim_{(x,y) \to (0,0)} g(x,y) = 0$. While it is true that along the $x$ and $y$ axis, we have

$$\lim_{x \to 0} g(x,0) = \lim_{y \to 0} g(0,y) = 0$$

if we approach the origin along the line $y = x$ we have

$$\lim_{x \to 0^{+}} g(x,x) = \lim_{x \to 0^{+}} \sin x^2 \frac{2x^3}{(x^2 + x^4)^2} = \lim_{x \to 0^{+}} \frac{\sin x^2}{x^2} \frac{2x^5}{(x^2 + x^4)^2} = \lim_{x \to 0^{+}} \frac{2x^5}{(x^2 + x^4)^2} = +\infty$$

so the limit does not exist and $f$ is not continuously differentiable at $(x,y) = (0,0)$.