How to prove that $f(x,y)=xy^2$ is differentiable everywhere via limit definition? 
How to prove that $f(x,y)=xy^2$ is differentiable everywhere via limit definition?

The thing is that I have to do this using the limit definition.
I arrived at this limit and I have not been able to go further.
$$\lim_{(x,y) \to (x_0, y_0)} \frac{xy^2 -x y^2_0 - 2yx_0 y_0 + 2x_0 y^2_0}{||(x-x_0, y-y_0)||}$$
This was obtained by calculating the gradient.
 A: First notice that for all point $(x_{0},y_{0})\in \mathbb{R}^{2}$, we have

*

*$f_{x}(x_{0},y_{0})=y_{0}^{2}$.

*$f_{y}(x_{0},y_{0})=2x_{0}y_{0}$.

Now $f$ is differentiable function at $(x_{0},y_{0})\in \mathbb{R}^{2}$ by definition (this definition is better in this case and is equivalent to the one you are using, see here in differentiability in higher dimensions)  if and only if,
$$\lim_{(h,k)\to (0,0)}\frac{f(x_0+h,y_0 +k)-f(x_{0},y_{0})-f_{x}(x_{0},y_{0})\cdot h-f_{y}(x_{0},y_{0})\cdot k}{\sqrt{h^{2}+k^{2}}}=0$$
Now, since $f(x,y)=xy^{2}$ we can re-write the limit above as
$$\lim_{(h,k)\to (0,0)}\frac{(x_{0}+h)(y_{0}+k)^{2}-x_{0}y_{0}^{2}-y_{0}^{2}h-2x_{0}y_{0}k}{\sqrt{h^{2}+k^{2}}}$$
But also we have in the numerator
$$(x_{0}+h)(y_{0}+k)^{2}-x_{0}y_{0}^{2}-y_{0}^{2}h-2x_{0}y_{0}k=x_{0}k^{2}+hk^{2}+2y_{0}kh$$
Setting tha change of variables to polar coordinates
$$\begin{cases}h=r\cos \theta,\\k=r\sin\theta\end{cases}$$ with $r\in\mathbb{R}^{+*}$ and $\theta\in [0,2\pi[$ so we have the limit is equivalent to
$$\lim_{r\to 0}\frac{x_{0}(r\sin \theta)^{2}+(r\cos\theta)(r\sin\theta)^{2}+2y_{0}(r\sin\theta)(r\cos\theta)}{r}=0$$
Therefore $f$ is differentiable function at $(x_{0},y_{0})$ and since the point is arbitrary point of $\mathbb{R}^{2}$ so $f$ is differentiable over all $\mathbb{R}^{2}$.
