multivariable calculus - How do I prove continuity of partial derivative of $f(x, y) = sqrt{x^4+y^4}$ at $(0,0)$?

Consider I have $$f(x, y) = \sqrt{x^4+y^4}$$
And I want to check if the function has partial derivatives continuous in point $$(x_0, y_0) = (0, 0)$$

I know theorem, that existence of continuous partial derivatives implies differentiability of this function, and differentiablity implies that function is continuous. I can also test this function for if it is differentiable. Can I use those information to check if partial derivatives are continuous? If not, what is standard method of calculating continuity of partial derivative?

$$f_x=\lim_{x\to0}{\frac{f(0+x, 0) - f(0, 0)}{x}} = 0$$

Answer

You have:
$$f_x(x,y)=
\begin{cases}
\frac{2x^3}{\sqrt{x^4+y^4}}&(x,y)\neq(0,0)\\
0&(x,y)=(0,0)\\
\end{cases}
$$
and this function is continuos in $\mathbb R^2$ because $|f_x(x,y)|\leq 2|x|$ for each $(x,y)\in\mathbb R^2$.
By simmetry, also $f_y$ is continuous in $\mathbb R^2$.