I have been given the following function:
$$f(x,y) = \begin{cases} \dfrac{xy(2x^2 - y^2)}{2x^2 + y^2} & \text{if $(x, y) \ne (0, 0)$} \\ 0 & \text{if $(x, y) = (0, 0)$} \end{cases}$$
Now we would like to show that the partial derivatives are both continuous at the point $(0, 0)$, but neither one of them is differentiable at (0, 0). First we should analyze the behavior at $(0, 0)$, as the quotient rule will not apply at that point, so we should go by the definition:
$$ f_x(0, 0) = \lim_{x \rightarrow 0} \dfrac{f(x, 0) - f(0, 0)}{x} = \lim_{x \rightarrow 0} \dfrac{0 - 0}{x} = 0 $$ $$ f_y(0, 0) = \lim_{y \rightarrow 0} \dfrac{f(0, y) - f(0, 0)}{y} = \lim_{y \rightarrow 0} \dfrac{0 - 0}{y} = 0$$
Then we can obtain the following partial derivatives:
$$ \frac{\partial f}{\partial x} = \begin{cases} \dfrac{4 x^4 y + 8 x^2 y^3 - y^5}{\left(2x^2 + y^2\right)^2} & \text{if $(x, y) \ne (0, 0)$} \\ 0 & \text{if $(x, y) = (0, 0)$} \end{cases}$$
$$ \frac{\partial f}{\partial y} = \begin{cases} \dfrac{4x^5 - 8x^3 y^2 - xy^4}{\left(2x^2 + y^2 \right)^2} & \text{if $(x, y) \ne (0, 0)$} \\ 0 & \text{if $(x, y) = (0, 0)$} \end{cases}$$
My goal is to show that both of them are continuous at $(0, 0)$, but neither is differentiable. I have already shown that both of them are continuous by showing that both partials have a limit of $0$ at $(0, 0)$. (I just took the limit in polar coordinates).
But to show differentiability, I am to use the following definition of differentiability:
A function $f(x, y)$ is differentiable at $(x_0, y_0)$ if it can be expressed in the form:
$$ f\left(x, y\right) = f\left(x_0, y_0\right) + \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \xi \left(x, y\right)$$
Where $\Delta x = x - x_0$, $\Delta y = y - y_0$, $r = \sqrt{(x - x_0)^2 + (y - y_0)^2}$ and $\xi$ has the following property:
For any $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $\delta < r$, $\left | \xi \right | < \epsilon r$.
In other words, $\left | \xi \right |$ disappears faster than any linear multiple of $r$ as the point is approached. I am to use this definition to prove that the partial derivatives are not differentiable. How could I prove that, for some $\epsilon > 0$, there is no valid choice of $\delta$?
Hint: You can see that $f_x(x,0)\equiv 0,$ so $f_{xx}(0,0) = 0.$ Also $f_x(0,y) = -y.$ Thus $f_{xy}(0,0) = -1.$ Now if $f_x$ were differentiable at $(0,0),$ we would have
$$f_x(x,y) = f_x(0,0) + f_{xx}(0,0)\cdot x + f_{xy}(0,0)\cdot y + \xi = -y + \xi.$$
Is this true? (Consider $f_x(x,x)$ for example.)