Proving continuous differentiability over the domain I have the following problem from Tao.

Let $f: \mathbb R^2 \to\mathbb R$ be the function defined by $f(x,y)=\dfrac {xy^3}{x^2+y^2}$ when $(x,y) \neq (0,0)$ and $f(0,0):=0$. Show that $f$ is continuously differerentiable.

I have $\dfrac{\partial{f(x,y)}}{\partial{x}} = \dfrac{y^5 - x^2y^3}{(x^2+y^2)^2}$ when $(x,y) \neq (0,0)$ and $\dfrac{\partial{f(x,y)}}{\partial{x}} = (0,0)$ when $(x,y) \neq (0,0)$.
I am not sure how to proceed with proving continuity, because it seems that the derivative is discontinuous. I am inclined to proceed with $\epsilon-\delta$ definition if possible, but I don't see the point.

Let $f(x,y)$ be given by
$$f(x,y)=\begin{cases}\frac{xy^3}{x^2+y^2}&,(x,y)\ne (0,0)\\\\0&,(x,y)=(0,0)\end{cases}$$
For $(x,y)\ne(0,0)$, we have
$$\begin{align}
\frac{\partial f(x,y)}{\partial  x}&=\frac{y^3(y^2-x^2)}{(x^2+y^2)^2}\\\\
\frac{\partial f(x,y)}{\partial  y}&=\frac{xy^2(3x^2+y^2)}{(x^2+y^2)^2}
\end{align}$$

Now, note that from the limit definition of the partial derivatives at $(0,0)$ we have
$$f_x(0,0)=f_y(0,0)=0$$

We now use the bounds $|x^2-y^2|\le x^2+y^2$, $|x|\le \sqrt{x^2+y^2}$ and $|y|\le \sqrt{x^2+y^2}$ to establish the estimates
$$\begin{align}
\left|\frac{\partial f(x,y)}{\partial x}\right|&=\left|\frac{y^3(y^2-x^2)}{(x^2+y^2)^2}\right|\\\\
&\le \frac{(x^2+y^2)^{5/2}}{(x^2+y^2)^2}\\\\
&=\sqrt{x^2+y^2}
\end{align}$$
Similarly, we have
$$\left|\frac{\partial f(x,y)}{\partial y}\right|=\left|\frac{x(3x^2y^2+y^4)}{(x^2+y^2)^2}\right|\le 3\sqrt{x^2+y^2}$$

Can you wrap things up now?
